Let $A\in M_{3\times 3}$ and $\vec x, \vec y, \vec z\in\mathbb R^3$. Suppose that we have $$A\vec x=\begin{vmatrix}1\\0\\1\end{vmatrix}, A\vec y=\begin{vmatrix}0\\1\\0\end{vmatrix}, A\vec z=\begin{vmatrix}1\\1\\1\end{vmatrix}.$$ Then find the value of the determinant of the matrix $A$.
Started by treating each equation as a row column in a $3 \times 3$ matrix and finding the determinant of that matrix. Not sure if on the right track.
In the comments, Al-Hasan Ibn Al-Hasan posted an idea that can be used to settle this question.
We were given that $A$ is $3\times3$, and that for some $\vec{x},\vec{y},\vec{z}\in\Bbb{R}^3$,
$$A\vec{x}=\begin{bmatrix}1\\0\\1\end{bmatrix},\quad A\vec{y}=\begin{bmatrix}0\\1\\0\end{bmatrix},\quad\text{and}\quad A\vec{z}=\begin{bmatrix}1\\1\\1\end{bmatrix}.$$
Using this information alone, there is absolutely nothing that we can conclude about the determinant of $A$. Let's look at an example to see why.
Let $\vec{x}=e_1$, let $\vec{y}=e_2$, and let $\vec{z}=e_1+e_2$. Let $\lambda$ be your favorite scalar. If we let
$$A=\begin{bmatrix}1&0&0\\0&1&0\\1&0&\lambda\end{bmatrix},$$
then we can check that $A$ satisfies the given conditions, and that $\det A=\lambda$.
It follows that if the only thing we know about $A$ is that it satisfies the given conditions, then $\det A$ can be any scalar. So the given conditions don't place any restrictions on the value of $\det A$.