Given $A=\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 3 & 2 \\ 3 & 4 & 5\end{array}\right]$
$$
\begin{aligned}
& \text { Now, } \mathrm{A}^{\mathrm{T}}=\left[\begin{array}{lll}
1 & 4 & 3 \\
2 & 3 & 4 \\
3 & 2 & 5
\end{array}\right] \\
& \left(\mathrm{A}+\mathrm{A}^{\mathrm{T}}\right)\left(\mathrm{A}-\mathrm{A}^{\mathrm{T}}\right)= \\
& \left(\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 3 & 2 \\
3 & 4 & 5
\end{array}\right]+\left[\begin{array}{ccc}
1 & 4 & 3 \\
2 & 3 & 4 \\
3 & 2 & 5
\end{array}\right]\right)\left(\left[\begin{array}{ccc}
1 & 2 & 3 \\
4 & 3 & 2 \\
3 & 4 & 5
\end{array}\right]-\left[\begin{array}{ccc}
1 & 4 & 3 \\
2 & 3 & 4 \\
3 & 2 & 5
\end{array}\right]\right) \\
& \left.=\left[\begin{array}{lll}
2 & 6 & 6 \\
6 & 6 & 6 \\
6 & 6 & 10
\end{array}\right]\left[\begin{array}{ccc}
0 & -2 & 0 \\
2 & 0 & -2 \\
0 & 2 & 0
\end{array}\right]\right) \\
& =\left[\begin{array}{ccc}
0+12+0 & -4+0+12 & 0-12+0 \\
0+12+0 & -12+0+12 & 0-12+0 \\
0+12+0 & -12+0+20 & 0-12+0
\end{array}\right] \\
& =\left[\begin{array}{lll}
12 & 8 & -12 \\
12 & 0 & -12 \\
12 & 8 & -12
\end{array}\right]=4\left[\begin{array}{ccc}
3 & 2 & -3 \\
3 & 0 & -3 \\
3 & 2 & -3
\end{array}\right]
\end{aligned}
$$
So, correct option is (a)