Answer
See below
Work Step by Step
Given: $A=\begin{bmatrix}
1 & 2 & 4\\
3 & 1 & 6\\
k & 3 & 2
\end{bmatrix}$
a) For $A$ to be invertible, the determinant should not be equal to zero.
Find $\det(A)=2+12k+36-4k-12-18=8+8k$
then $\det(A)=8k+8 \ne 0 \rightarrow k\ne -1$
b) In terms of $k$, we will determine the volume of the parallelepiped determined by the row vectors of the matrix $A$
$V_1=\begin{vmatrix}
1 & 3& k\\
2 & 1& 3\\
4 & 6 & 2
\end{vmatrix}=8+8k$
and we have $V_2=\det(A)=8+8k$
We can see that $V_1=V_2$.
Without calculation, we still can know the volumes are the same because according to Property P4, we know $V_1=|A^T|,|A|=|A^T|$
