\[
\text { Let } M \text { be a } 3 \times 3 \text { matrix satisfying }
\]

\[
M\left[\begin{array}{l}
0 \\
1 \\
0
\end{array}\right]=\left[\begin{array}{c}
-1 \\
2 \\
3
\end{array}\right], M\left[\begin{array}{c}
1 \\
-1 \\
0
\end{array}\right]=\left[\begin{array}{c}
1 \\
1 \\
-1
\end{array}\right] \text { and } M\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right]=\left[\begin{array}{c}
0 \\
0 \\
12
\end{array}\right]
\]
Then, the sum of the diagonal entries of \(M\) is

Let \(M=\left[\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right]\)
\(\therefore M\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]=\left[\begin{array}{r}-1 \\ 2 \\ 3\end{array}\right], M\left[\begin{array}{r}1 \\ -1 \\ 0\end{array}\right]=\left[\begin{array}{r}1 \\ 1 \\ -1\end{array}\right]\),
\(M\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[\begin{array}{r}0 \\ 0 \\ 12\end{array}\right]\),
\[
\begin{aligned}
\Rightarrow & {\left[\begin{array}{l}
a_2 \\
b_2 \\
c_2
\end{array}\right]=\left[\begin{array}{r}
-1 \\
2 \\
3
\end{array}\right],\left[\begin{array}{l}
a_1-a_2 \\
b_1-b_2 \\
c_1-c_2
\end{array}\right]=\left[\begin{array}{r}
1 \\
1 \\
-1
\end{array}\right], } \\
& {\left[\begin{array}{l}
a_1+a_2+a_3 \\
b_1+b_2+b_3 \\
c_1+c_2+c_3
\end{array}\right]=\left[\begin{array}{r}
0 \\
0 \\
12
\end{array}\right] } \\
\Rightarrow & a_2=-1, b_2=2, c_2=3, a_1-a_2=1, \\
\Rightarrow & b_1-b_2=1, c_1-c_2=-1 \\
\Rightarrow \quad & a_1+a_2+a_3=0, b_1+b_2+b_3=0, \\
\therefore & c_1+c_2+c_3=12 \\
\therefore & a_1=0, b_2=2 \text { and } c_3=7
\end{aligned}
\]
Hence, sum of diagonal elements
\[
=0+2+7=9
\]