Answer
The matrix $\left( A-D \right)C $ is, $\left( A-D \right)C=\left[ \begin{array}{*{35}{l}}
7 & 6 & 5 \\
2 & -1 & 11 \\
\end{array} \right]$.
Work Step by Step
Here we need to find $\left( A-D \right)C $. Therefore consider, $\begin{align}
& \left( A-D \right)C=\left( \left[ \begin{array}{*{35}{l}}
2 & -1 & 2 \\
5 & 3 & -1 \\
\end{array} \right]-\left[ \begin{array}{*{35}{l}}
-2 & 3 & 1 \\
3 & -2 & 4 \\
\end{array} \right] \right)\left[ \begin{matrix}
1 & 2 & 3 \\
-1 & 1 & 2 \\
-1 & 2 & 1 \\
\end{matrix} \right] \\
& =\left( \left[ \begin{array}{*{35}{l}}
2+2 & -1-3 & 2-1 \\
5-3 & 3+2 & -1-4 \\
\end{array} \right] \right)\left[ \begin{matrix}
1 & 2 & 3 \\
-1 & 1 & 2 \\
-1 & 2 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{array}{*{35}{l}}
4 & -4 & 1 \\
2 & 5 & -5 \\
\end{array} \right]\left[ \begin{matrix}
1 & 2 & 3 \\
-1 & 1 & 2 \\
-1 & 2 & 1 \\
\end{matrix} \right]
\end{align}$
It can be further simplified as:
$\begin{align}
& \left[ \begin{array}{*{35}{l}}
4 & -4 & 1 \\
2 & 5 & -5 \\
\end{array} \right]\left[ \begin{matrix}
1 & 2 & 3 \\
-1 & 1 & 2 \\
-1 & 2 & 1 \\
\end{matrix} \right]=\left[ \begin{array}{*{35}{l}}
4\left( 1 \right)-4\left( -1 \right)+1\left( -1 \right) & 4\left( 2 \right)-4\left( 1 \right)+1\left( 2 \right) & 4\left( 3 \right)-4\left( 2 \right)+1\left( 1 \right) \\
2\left( 1 \right)+5\left( -1 \right)-5\left( -1 \right) & 2\left( 2 \right)+5\left( 1 \right)-5\left( 2 \right) & 2\left( 3 \right)+5\left( 2 \right)-5\left( 1 \right) \\
\end{array} \right] \\
& =\left[ \begin{array}{*{35}{l}}
4+4-1 & 8-4+2 & 12-8+1 \\
2-5+5 & 4+5-10 & 6+10-5 \\
\end{array} \right] \\
& =\left[ \begin{array}{*{35}{l}}
7 & 6 & 5 \\
2 & -1 & 11 \\
\end{array} \right]
\end{align}$
Thus, $\left( A-D \right)C=\left[ \begin{array}{*{35}{l}}
7 & 6 & 5 \\
2 & -1 & 11 \\
\end{array} \right]$