Answer
See below
Work Step by Step
Given: $A=\begin{bmatrix}
k+1 & 2 & 1\\
0 & 3 & k\\
1 & 1 & 1
\end{bmatrix}$
a) For $A$ to be invertible, the determinant should not be equal to zero.
Find $\det(A)=3(k+1)+2k-3-k(k+1)=3+2k-k^2+2k-3=4k-k^2$
then $\det(A)=4k-k^2 \ne 0 \rightarrow k\ne 0, k\ne 4$
b) In terms of $k$, we will determine the volume of the parallelepiped determined by the row vectors of the matrix $A$
$V_1=\begin{vmatrix}
k+1& 0& 1\\
2 & 3& 1\\
1 & k & 1
\end{vmatrix}=4k-k^2$
and we have $V_2=\det(A)=4k-k^2$
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 $V_1=|A^T|,|A|=|A^T|$
