The method at that site ignores the sign of the determinant, and is phrased in a way that makes the generalization to higher dimensions a bit less clear than I'd like. I'll try here to provide an altered version of the approach up to four dimensions. I'll use bold for vectors, subscripts for components, and double vertical lines for length, so that $\left\Vert \mathbf{a}\right\Vert =\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots}$.
As was mentioned by LutzL in a comment, this method is very closely connected to using the $QR$-decomposition to find the absolute value of the determinant, as described on Wikipedia here.
1D:
Let's calculate $\det\left(\mathbf{a}\right)$ where $\mathbf{a}$ has one nonzero component. It's $\left\Vert \mathbf{a}\right\Vert$ if $\mathbf{a}$ has a positive component and $-\left\Vert \mathbf{a}\right\Vert$ if $\mathbf{a}$ has a negative component.
2D:
Let's calculate $\det\left(\mathbf{a},\mathbf{b}\right)$ where $\mathbf{a}$ and $\mathbf{b}$ are not collinear.
Let's ignore $\mathbf{a}$ for now. The first step is to find a vector $\mathbf{n}$ that's orthogonal to $\mathbf{b}$. We set $\mathbf{n}\bullet\mathbf{b}$ equal to $\boldsymbol{0}$. That's two unknowns and only one equation. In a typical case, the component $n_{1}$ of $\mathbf{n}$ is not forced to be $0$, so it can be whatever we want that's nonzero (e.g. $1$). (In a special case, $n_{1}$ might be forced to be $0$, but then $n_{2}$ can be chosen freely.)
Now scale $\mathbf{n}$ to get a new vector $\mathbf{n}'$ so that $\left\Vert \mathbf{n}'\right\Vert =\left\Vert \mathbf{b}\right\Vert$ . By some geometry, the area of the parallelogram formed by $\mathbf{a}$ and $\mathbf{b}$ is then $\left|\mathbf{n}'\bullet\mathbf{a}\right|$. The determinant is $\pm\mathbf{n}'\bullet\mathbf{a}$ where the $\pm$ sign here (which may not be the sign of the determinant) is positive exactly when the rotation to get from $\mathbf{a}$ to $\mathbf{b}$ is in the same direction (clockwise or counterclockwise) as the rotation to get from $\mathbf{b}$ to $\mathbf{n}'$. Unfortunately, that can't be determined by a dot-product calculation.
3D:
Let's calculate $\det\left(\mathbf{a},\mathbf{b},\mathbf{c}\right)$ where $\mathbf{a},\mathbf{b},\mathbf{c}$ are not coplanar.
Let's ignore $\mathbf{a}$ for now. The first step is to find a vector $\mathbf{n}$ that's orthogonal to both $\mathbf{b}$ and $\mathbf{c}$. We set $\mathbf{n}\bullet\mathbf{b},\mathbf{n}\bullet\mathbf{c}$ equal to $\boldsymbol{0}$. That's three unknowns and only two equations. In a typical case, the component $n_{1}$ of $\mathbf{n}$ is not forced to be $0$, so it can be whatever we want that's nonzero (e.g. $1$). (In a special case, $n_{1}$ might be forced to be $0$, but then $n_{2}$ or $n_{3}$ can be chosen freely.)
The second step is to find a vector $\mathbf{o}$ that's orthogonal to $\mathbf{c}$ (this choice differs from the original author), but lies in the same plane as $\mathbf{b}$ and $\mathbf{c}$. To keep it in that plane, we need $\mathbf{o}$ to be orthogonal to $\mathbf{n}$. So we have $\mathbf{o}\bullet\mathbf{n}=\boldsymbol{0}$ as well as $\mathbf{o}\bullet\mathbf{c}=\boldsymbol{0}$. Again that's three unknowns and two equations, so we have a degree of freedom and could choose a particular value for some component.
Now scale $\mathbf{o}$ to get a new vector $\mathbf{o}'$ so that $\left\Vert \mathbf{o}'\right\Vert =\left\Vert \mathbf{c}\right\Vert$ . By some geometry, the area of the parallelogram formed by $\mathbf{b}$ and $\mathbf{c}$ is then $\left|\mathbf{o}'\bullet\mathbf{b}\right|$. Now scale $\mathbf{n}$ to get a new vector $\mathbf{n}'$ so that $\left\Vert \mathbf{n}'\right\Vert =\left|\mathbf{o}'\bullet\mathbf{b}\right|$. By some geometry, the volume of the parallelepiped formed by $\mathbf{a}, \mathbf{b}, \mathbf{c}$ is then $\left|\mathbf{n}'\bullet\mathbf{a}\right|$. The determinant is then $\pm\mathbf{n}'\bullet\mathbf{a}$ where I'm pretty sure the $\pm$ sign is positive when the handedness (right or left) of $\mathbf{a},\mathbf{b},\mathbf{c}$ is the same as that of $\mathbf{c},\mathbf{n}',\mathbf{o}'$.
4D:
Let's calculate $\det\left(\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}\right)$ where $\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}$ are not in the same $3$-dimensional hyperplane.
Let's ignore $\mathbf{a}$ for now. The first step is to find a vector $\mathbf{n}$ that's orthogonal to all three of $\mathbf{b},\mathbf{c},\mathbf{d}$. We set $\mathbf{n}\bullet\mathbf{b},\mathbf{n}\bullet\mathbf{c},\mathbf{n}\bullet\mathbf{c}$ all to $\boldsymbol{0}$. That's four unknowns and only three equations. In a typical case the component $n_{1}$ of $\mathbf{n}$ is not forced to be $0$, so it can be whatever we want that's nonzero (e.g. $1$). (In a special case, $n_{1}$ might be forced to be $0$, but there will be at least one component we could choose freely.)
The second step is to find a vector $\mathbf{o}$ that's orthogonal to $\mathbf{c}$ and $\mathbf{d}$, but lies in the same $3$-dimensional hyperplane as $\mathbf{b},\mathbf{c},\mathbf{d}$. To keep it in that hyperplane, we need $\mathbf{o}$ to be orthogonal to $\mathbf{n}$. So we have $\mathbf{o}\bullet\mathbf{n}=\boldsymbol{0}$ as well as $\mathbf{o}\bullet\mathbf{c},\mathbf{o}\bullet\mathbf{d}=\boldsymbol{0}$. Again that's four unknowns and three equations, so we have a degree of freedom and could choose a particular value for some component.
The third step is to find a vector $\mathbf{p}$ that's orthogonal to $\mathbf{d}$, but lies in the same $2$-dimensional plane as $\mathbf{c},\mathbf{d}$. To keep it in that plane, it should be orthogonal to $\mathbf{n}$ as well as $\mathbf{o}$. So we have $\mathbf{p}\bullet\mathbf{n},\mathbf{p}\bullet\mathbf{o}=\boldsymbol{0}$ as well as $\mathbf{p}\bullet\mathbf{d}=\boldsymbol{0}$. Again that's four unknowns and three equations, so we have a degree of freedom and could choose a particular value for some component.
Now scale $\mathbf{p}$ to get a new vector $\mathbf{p}'$ so that $\left\Vert \mathbf{p}'\right\Vert =\left\Vert \mathbf{d}\right\Vert$ . By some geometry, the area of the parallelogram formed by $\mathbf{c}$ and $\mathbf{d}$ is then $\left|\mathbf{p}'\bullet\mathbf{c}\right|$. Now scale $\mathbf{o}$ to get a new vector $\mathbf{o}'$ so that $\left\Vert \mathbf{o}'\right\Vert =\left|\mathbf{p}'\bullet\mathbf{c}\right|$. By some geometry, the volume of the parallelepiped formed by $\mathbf{b}, \mathbf{c}, \mathbf{d}$ is then $\left|\mathbf{o}'\bullet\mathbf{b}\right|$. Now scale $\mathbf{n}$ to get a new vector $\mathbf{n}'$ so that $\left\Vert \mathbf{n}'\right\Vert =\left|\mathbf{o}'\bullet\mathbf{b}\right|$. By some geometry, the hypervolume of the hyperparallelepiped formed by $\mathbf{a},\mathbf{b}, \mathbf{c}, \mathbf{d}$ is then $\left|\mathbf{n}'\bullet\mathbf{a}\right|$. The determinant is then $\pm\mathbf{n}'\bullet\mathbf{a}$ where I think the $\pm$ sign is positive when the orientation of $\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}$ is the same as that of $\mathbf{d},\mathbf{n}',\mathbf{o}',\mathbf{p}'$.
Best Answer
The problem statement seems to assume that the displacements are as good as infinitesimal; I will therefore omit some differential symbols and assume that the sled's motion and the ground are approximately uncurved in this problem. "Motion" here means displacement as a function of time.
The second question in the problem statement suggests that different forces $\vec{F}$ can have exactly the same effect on the motion of the sled. In other words, there are nonzero force changes corresponding to zero motion changes. This implies that the system is constrained.
The constraint that comes to mind for a typical sled on the ground is that it only follows the direction of its skids. Since we are assuming that the geometry here is as good as infinitesimal, this can be simplified to mean that the sled moves only along a fixed axis. Obviously, the axis determines the direction (up to sign) of $\vec{s}$.
It will be helpful to decompose $$\vec{F} = \vec{F}_s + \vec{F}_o$$ where $\vec{F}_s$ is collinear with $s$ and $\vec{F}_o$ is orthogonal to $\vec{s}$. Thus, $\vec{F}_s$ is your projection of $\vec{F}$ on $\vec{s}$. You can achieve this with the following matrix $\mathbf{S}$: $$\begin{align} \text{Let}\quad \mathbf{S} &= \frac{1}{\vec{s}\cdot\vec{s}}\,\vec{s}\circ\vec{s} = \frac{1}{35}\begin{pmatrix}5\\-1\\3\end{pmatrix} \begin{pmatrix}5&-1&3\end{pmatrix} \\\text{Then}\quad \vec{F}_s &= \mathbf{S}\cdot\vec{F} = \frac{\vec{F}\cdot\vec{s}}{\vec{s}\cdot\vec{s}}\,\vec{s} ,\quad\vec{F}_o = \vec{F}-\vec{F}_s = (\mathbf{I}-\mathbf{S})\cdot\vec{F} \end{align}$$ where $\cdot$ denotes the dot product, $\circ$ denotes the outer product, and $\mathbf{I}$ is the $3\times3$ identity matrix. Note that $\mathbf{S}$ is symmetric and positive semidefinite. Furthermore, $\mathbf{S}$ is idempotent: $\mathbf{S}\cdot\mathbf{S}=\mathbf{S}$. This allows you to verify that $\vec{F}_s\cdot\vec{F}_o=0$.
Back to the mechanics. The balance of forces requires that all force components orthogonal to the sled's acceleration sum to zero. Due to the constraint imposed, acceleration can happen only along the axis determined by $\vec{s}$, therefore $\vec{F}_o$ must be counteracted by a constraint force $$\vec{F}_c=-\vec{F}_o$$
The fact that $\vec{F}_o\cdot\vec{s}=0$ also means that $\vec{F}_o$ does not contribute to the work done. The same applies to $\vec{F}_c$ and is consistent with the general fact that constraint forces do not do work: $$W = \vec{F}\cdot\vec{s} = \vec{F}_s\cdot\vec{s} + \underbrace{\vec{F}_o\cdot\vec{s}}_0 = \vec{F}_s\cdot\vec{s}$$ In fact, the sled's motion can only be influenced by the sum $$\vec{F}+\vec{F}_c = \vec{F}_s+ \underbrace{\vec{F}_o+\vec{F}_c}_0 = \vec{F}_s$$ which again points to the conclusion that preserving motion requires preserving $\vec{F}_s$, whereas $\vec{F}_o$ can be varied arbitrarily without effect on the sled's motion.
Now, since $\vec{F}_s$ and $\vec{F}_o$ are orthogonal by design, we can apply the pythagorean theorem: $$\|\vec{F}\|^2 = \|\vec{F}_s\|^2+\|\vec{F}_o\|^2$$
And this means that minimization of $\|\vec{F}\|$ while preserving $\vec{F}_s$ means minimizing $\|\vec{F}_o\|$. This is achieved with $\vec{F}_o=\vec{0}$. Thus, the minimal required force is $$\vec{F} = \vec{F}_s = \frac{\vec{F}\cdot\vec{s}}{\vec{s}\cdot\vec{s}}\,\vec{s} = \frac{26\,\mathrm{Nm}}{35\,\mathrm{m}^2}\vec{s} \quad\text{with}\quad \|\vec{F}\| = \|\vec{F}_s\| = \frac{26}{\sqrt{35}}\mathrm{N}$$