Let
$$
\vec{r}(u,v) = \left [ \begin{matrix} x(u,v) \\ y(u,v) \\ z(u,v) \end{matrix} \right ], \quad
\vec{r}_u (u,v) = \left [ \begin{matrix} x_u (u,v) \\ y_u (u,v) \\ z_u (u,v) \end{matrix} \right ], \quad
\vec{r}_v (u,v) = \left [ \begin{matrix} x_v (u,v) \\ y_v (u,v) \\ z_v (u,v) \end{matrix} \right ]
$$
where
$$
\begin{aligned}
x_u (u,v) &= \frac{d\,x(u,v)}{d\,u}, \\
y_u (u,v) &= \frac{d\,y(u,v)}{d\,u}, \\
z_u (u,v) &= \frac{d\,z(u,v)}{d\,u}, \\
\end{aligned} \quad \begin{aligned}
x_v (u,v) &= \frac{d\,x(u,v)}{d\,v} \\
y_v (u,v) &= \frac{d\,y(u,v)}{d\,v} \\
z_v (u,v) &= \frac{d\,z(u,v)}{d\,v} \\
\end{aligned}
$$
then
$$
\vec{n} (u, v) = \vec{r}_u (u,v) \times \vec{r}_v (u,v) = \left [ \begin{matrix}
y_u (u,v) \, z_v (u,v) - y_v (u,v) \, z_u (u,v) \\
z_u (u,v) \, x_v (u,v) - z_v (u,v) \, x_u (u,v) \\
x_u (u,v) \, y_v (u,v) - x_v (u,v) \, y_u (u,v) \\
\end{matrix} \right ]
$$
Rather than answer the stated question, I'll try to show how to find the answer to any similar problem instead.
$\vec{r}(u, v)$ describes a surface.
$\vec{r}_u (u,v)$ and $\vec{r}_v (u,v)$ are the tangent vectors at $(u,v)$. They form the surface tangent plane at $(u,v)$. They are not necessarily perpendicular to each other; it depends on the surface.
$\vec{n} (u,v) = \vec{r}_u (u,v) \times \vec{r}_v (u,v)$ is the surface normal at $(u, v)$. This vector is always perpendicular to $\vec{r}_u (u,v)$ and $\vec{r}_v (u,v)$ by definition.
Surface unit normal is just the surface normal scaled to length 1:
$$\hat{n} (u,v) = \frac{\vec{n} (u,v)}{\lVert \vec{n}(u,v) \rVert}$$
If $\lVert\vec{r}(u, v)\rVert$ is a constant, it means the points on the surface are at a fixed distance from origin; thus, the surface is (part of) a spherical shell.
In a sphere, the surface normal is always parallel to the radius vector. Thus, if $\lVert\vec{r}(u, v)\rVert = R$, then $R \hat{n}(u,v) = \vec{r}(u, v)$. Obviously, if $\lVert\vec{r}(u,v)\rVert = 1$, then $\hat{n}(u, v) = \vec{r}(u, v)$.
Although the direction of the tangent vectors is dictated by the surface shape, their length depends on the parametrization of the surface. Thus, the length of the surface normal vector $\vec{n}(u,v)$ also depends on how the surface is parametrized.
For example, using the typical parametrization for the unit sphere,
$$
\left\lbrace\begin{aligned}
x(u, v) &= \sin(u) \cos(v) \\
y(u, v) &= \sin(u) \sin(v) \\
z(u, v) &= \cos(u) \\
\end{aligned}\right.
$$
you get $\lVert\vec{r}_u (u,v) \rVert = 1$, but $\lVert\vec{r}_v (u,v) \rVert = \lvert \sin(u) \rvert$ and $\lVert\vec{n}(u,v)\rVert = \lvert\sin(u)\rvert$. This is because of the poles in this parametrization.
We can parametrize the positive $z$ half of the unit sphere using
$$\left\lbrace\begin{aligned}
x(u,v) &= u \\
y(u,v) &= v \\
z(u,v) &= \sqrt{1 - u^2 - v^2} \\
\end{aligned}\right.$$
In this case, the pole is absolutely fine, but the circle at $z = 0$ is problematic. The lengths of the tangents are $\lVert\vec{r}_u (u,v)\rVert = \frac{1 - v^1}{\sqrt{1-u^2-v^2}}$ and $\lVert\vec{r}_v (u,v)\rVert = \frac{1 - u^2}{\sqrt{1-u^2-v^2}}$; the length of the normal is $\lVert\vec{n}(u,v)\rVert = \frac{1}{\sqrt{1-u^2-v^2}}$.
For ease of typing let's use $\{x,y,z\}$ in place of $\{r_1,r_2,r_3\}$.
And $(a,b)$ for the differences instead of $\{(z-x),\,(y-x)\}$.
The vector we wish to analyze is $(a\times b)$.
$$\eqalign{
p &= a\times b &\implies dp = a\times db - b\times da \\
\lambda^2 &= p\cdot p &\implies \lambda\,d\lambda = p\cdot dp,\;
\lambda=\|p\| \\
n &= \lambda^{-1}p &\implies d\lambda=n\cdot dp,\quad
{\tt 1}=\|n\| \\
}$$
Dot an arbitrary vector with $dp$ and recall the triple scalar product rule:
$\;c\cdot(a\times b) = (c\times a)\cdot b$
$$\eqalign{
y\cdot dp
&= y\cdot(a\times db) - y\cdot(b\times da) \\
&= (y\times a)\cdot db - (y\times b)\cdot da \\
&= (Ya)\cdot db - (Yb)\cdot da \\
&= Y\,(a\cdot db - b\cdot da) \\
}$$
where the uppercase letter $\{Y\}$ denotes the skew symmetric cross-product matrix for $y$.
Now we're ready to tackle the derivative of the ${\cal E}$ function.
$$\eqalign{
{\cal E} &= v\cdot n = \lambda^{-1}\;v\cdot p \\
d{\cal E} &= \lambda^{-1}\;v\cdot dp - \lambda^{-2}d\lambda\;v\cdot p \\
&= \lambda^{-1}\;(v\cdot dp - {\cal E}\,d\lambda) \\
&= \lambda^{-1}\;(v\cdot dp-{\cal E}n\cdot dp) \\
&= \lambda^{-1}\;(v-{\cal E}n)\cdot dp \\
&= \lambda^{-1}\;(V-{\cal E}N)\,(a\cdot db - b\cdot da) \\
}$$
If $\{y,z\}$ are held constant we can substitute $\;da=db=-dx\;$
$$\eqalign{
d{\cal E}
&= \lambda^{-1}\,(V-{\cal E}N)(b-a)\cdot dx \\
&= \lambda^{-1}\,(v-{\cal E}n)\times(y-z)\cdot dx \\
\frac{\partial{\cal E}}{\partial x}
&= \lambda^{-1}\,(v-{\cal E}n)\times(y-z) \\
}$$
Conversely, if we hold $\{x,z\}$ constant, then $\,da=0,\,db=dy$.
While holding $\{x,y\}$ constant means $\;da=dz,\,db=0$.
Best Answer
When we write a cross-product as a determinant, we are really abusing notation. If you look at the expression, \begin{align} \mathbf{a\times b} &= \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ a_1&a_2&a_3\\ b_1&b_2&b_3\\ \end{vmatrix} \\ &= (a_2b_3 - a_3b_2)\mathbf{i} -(a_1b_3 - a_3b_1)\mathbf{j} +(a_1b_2 - a_2b_1)\mathbf{k} \\ &=\begin{pmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{pmatrix}, \end{align} you can see that in each term we have a product of two dimensionful `length' coordinates, giving an interpretation as area.
When we talk about it being numerically the area of a parallogram, we are talking about the norm of this vector, \begin{align} \left\lVert \mathbf{a\times b} \right\rVert &= \sqrt{(a_2b_3 - a_3b_2)^2 + (a_1b_3 - a_3b_1)^2 + (a_1b_2 - a_2b_1)^2}. \end{align} Counting the dimensions is easiest if we just consider \begin{align} &\mathbf{a} = \begin{pmatrix} a_1 \\ 0 \\ 0\end{pmatrix} &\mathbf{b} = \begin{pmatrix} 0 \\ b_2 \\ 0\end{pmatrix}, \end{align} then \begin{align} \left\lVert \mathbf{a\times b} \right\rVert &= \sqrt{(a_1b_2)^2} \\ &= a_1 b_2, \end{align} ie an area.
Why do we understand a determinant as corresponding to volume? Consider instead the scalar triple product, \begin{align} \left(\mathbf{a\times b}\right) \cdot \mathbf{c} &= \left( a_2b_3-a_3b_2 \right) c_1 + \left( a_3b_1-a_1b_3 \right) c_2 + \left( a_1b_2-a_2b_1 \right) c_3 \\ &= \begin{vmatrix} c_1&c_2&c_3\\ a_1&a_2&a_3\\ b_1&b_2&b_3\\ \end{vmatrix}. \end{align}
Then if we count the dimensions, eg by choosing \begin{align} &\mathbf{a} = \begin{pmatrix} a_1 \\ 0 \\ 0\end{pmatrix} &\mathbf{b} &= \begin{pmatrix} 0 \\ b_2 \\ 0\end{pmatrix} &\mathbf{c} = \begin{pmatrix} 0 \\ 0 \\ c_3\end{pmatrix}, \end{align} to simplify the algebra, we get $$ \left(\mathbf{a\times b}\right) \cdot \mathbf{c} = a_1 b_2 c_3, $$ a product of three lengths and hence a volume.
Final comment. Why do we abuse notation to write a cross-product as a determinant? Really, we are interested in the only totally-antisymmetric tensor in three dimensions, called the Levi-Civita, or alternating, symbol $\varepsilon_{ijk}$, defined as $$ \varepsilon_{ijk} = \begin{cases} +1 & \text{if } (i,j,k) \text{ is } (1,2,3), (2,3,1), \text{ or } (3,1,2), \\ -1 & \text{if } (i,j,k) \text{ is } (3,2,1), (1,3,2), \text{ or } (2,1,3), \\ \;\;\,0 & \text{if } i = j, \text{ or } j = k, \text{ or } k = i \end{cases}. $$
Both the cross-product and determinant are properly defined using this, with $$ \mathbf{a}\times\mathbf{b} = \sum_{i,j,k=1}^3 \mathbf{e}_i \;\varepsilon_{ijk} \, a_j b_k $$ and $$ \det A = \begin{vmatrix} A_{11}&A_{12}&A_{13}\\ A_{21}&A_{22}&A_{23}\\ A_{31}&A_{32}&A_{33}\\ \end{vmatrix} = \sum_{i,j,k=1}^3 \varepsilon_{ijk} \, A_{1i} A_{2j} A_{3k} $$ which allows us to use our mnemonic for calculating the latter (an expansion in the minors, the $2\times 2$ submatrices) to remember how to calculate the former.