The question:
solve the equation
$8x + 3y - 7z = 12 \\ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -\dfrac 3 8 \\ -\dfrac 8 3 \\ \dfrac 8 7 \end{bmatrix} + s \begin{bmatrix} \dfrac 7 8 \\ \dots \\ \dots \end{bmatrix} + t \begin{bmatrix} \dfrac 3 2 \\ \dots \\ \dots \end{bmatrix}$
I do not understand what to put into the remaining values. I tried to solve for the $y$ and $z$ like I did for the $x$, but the system is telling me that is incorrect. Some help would be appreciated.
Your equation is $8x+3y-7z=12$, hence $\displaystyle x=\frac{12-3y+7z}{8}=\frac{3}{2}-\frac{3}{8}y+\frac{7}{8}z$, thus the set of solutions is $\displaystyle S=\left\{\begin{pmatrix}\frac{3}{2}-\frac{3}{8}t+\frac{7}{8}s \\t\\s\end{pmatrix}\right\}$ or equivalently $\displaystyle \begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}\frac{3}{2}\\0\\0\end{pmatrix}+t\begin{pmatrix}-\frac{3}{8}\\1\\0\end{pmatrix}+s\begin{pmatrix}\frac{7}{8}\\0\\1\end{pmatrix}$.