(I edit my text for providing a solution closer to what you are looking for).
The fact that you give the general solution of the original system has no correlation with the question "solve the corresponding homogeneous system".
[In fact, it will be important to have both for an inevitable second question which is "deduce the general solution of the homogeneous system". See last line of this answer].
Some keypoints:
a) Consider the issue as looking for the kernel of a $3 \times 4$ matrix $A$ which acts as a linear operator with source space $\mathbb{R}^4$ and range space $\mathbb{R}^3$.
b) Minor the dimension of the range space $dim(Im(A)) \geq 2$ because the first two columns of $A$ are independant.
c) Use the rank-nullity theorem:
$dim(Ker(A))+dim(Im(A)=dim(source \ space))=4$.
Thus $dim(Ker(A)) \leq 2$ and in fact, we are able to exhibit two independent vectors of the kernel (by looking for null linear combinations of the columns of $A$, by trial and error for example, which usually does not take a long time when the coefficients are integers or almost integers as is the case here. The coefficients of the null combinations are naturaly the coordinates of elements of the kernel).
One of the elements of the kernel is $(10,-6,1,0)$, another one, non proportional to the first, is $(0,3,2,-5)$ ; thus we have a basis of the kernel.
As a consequence, a final solution to your question is
$(w,x,y,z)=a(10,-6,1,0)+b(0,3,2,-5)$ for any $a,b$.
Remark: this solution could appear under very different forms according to the basis chosen for $Ker(A)$.
and, consequently, the general solution is
$(w,x,y,z)=a(10,-6,1,0)+b(0,3,2,-5)+(0,2,3,-1)$ for any $a,b$.
You have here two of the fundamental ways to represent a line in $\mathbb R^2$. The first is described by a parametric representation that uses a point $\mathbf p_0$ on the line and a direction vector $\mathbf v$ parallel to the line. Every point on the line can then be expressed in the form $\mathbf p_0+\lambda\mathbf v$, $\lambda\in\mathbb R$.
The second line is described implicitly by an equation in point-normal form. There are various ways to interpret it geometrically, but I find this one easiest to visualize: The perpendicular line through the origin to a line $\mathscr l$ is completely characterized by a direction vector $\mathbf n$. Per the preceding paragraph, a parameterization of that perpendicular line is simply $\lambda\mathbf n$. If $\mathscr l$ is translated so that it passes through the origin, the translated line $\mathscr l'$ has the property that the position vector $\mathbf x'$ of every point on it is orthogonal to $\mathbf n$. That is, $\mathscr l'$ is the set of points that satisfy $\mathbf n\cdot\mathbf x'=0$. We can translate $\mathscr l$ in this way by fixing some point $\mathbf p$ on it and subtracting this from every point $\mathbf x$ on $\mathscr l$. Substituting into the orthogonality condition, an equation for $\mathscr l$ is therefore $\mathbf n\cdot(\mathbf x-\mathbf p)=0$, or $\mathbf n\cdot\mathbf x=\mathbf n\cdot\mathbf p$. Observe that it doesn’t matter which point was chosen for $\mathbf p$—the last form of the point-normal equation says that the dot product of any point on $\mathscr l$ with the normal vector $\mathbf n$ is constant.
Multiplying both sides of that last equation by $\mathbf n/\|\mathbf n\|^2$, we have $${\mathbf n \cdot \mathbf x\over \mathbf n\cdot\mathbf n}\mathbf n = {\mathbf n \cdot \mathbf p\over \mathbf n\cdot\mathbf n}\mathbf n,$$ which says that every point of $\mathscr l$ has the same orthogonal projection onto the normal vector $\mathbf n$. The length of this projection is $|\mathbf n\cdot\mathbf p|/\|\mathbf n\|$, which is also the distance of the line from the origin. Thus, the constant term in the point-normal equation of a line is the distance to the origin multiplied by the length of the normal vector. The sign of the constant term tells you on which side of the origin relative to the direction $\mathbf n$ the line lies.
To convert from the point-direction form to point-normal form, you have to find a normal to the line, which is any vector perpendicular to the line’s direction vector, and vice-versa when converting from point-normal to point-direction. The constant term for the point-normal form is then found by computing the dot product of this normal vector with any point on the line, as described above.
It’s not really necessary to convert representations to solve this problem, though. The intersection of the two lines is a point on the first line that satisfies the equation of the second, so you could simply plug the generic expressions for $x$ and $y$ from the first into the second and solve for $\lambda$ without doing any conversion. I expect that your professor has you do the conversion because he wants you to practice solving systems of linear equations, which is a major motivation for linear algebra in the first place.
Best Answer
Yes $x,y$ are scalars. Yes the column vectors are the same as you mentioned. This notation is classical in linear algebra because you can go ahead and write the above as $$\begin{bmatrix} 1 & -4 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ 6 \end{bmatrix}$$ where now all you have to do is "invert" the matrix to obtain $x,y$.