Let $\mathrm{A}=\left[\begin{array}{cc}3 x+7 & 5 \\ y+1 & 2-3 x\end{array}\right]=\left[\begin{array}{cc}0 & y-2 \\ 8 & 4\end{array}\right]$
(a) For $x=\frac{-1}{3}, y=7$
$$
\begin{aligned}
&\text { Now }\left[\begin{array}{cc}
3 \times\left(\frac{-1}{3}\right) & 5 \\
7+1 & 2-3 \times\left(\frac{-1}{3}\right)
\end{array}\right]=\left[\begin{array}{cc}
0 & 7-2 \\
8 & 4
\end{array}\right] \\
&\Rightarrow\left[\begin{array}{cc}
-1 & 5 \\
8 & 3
\end{array}\right]=\left[\begin{array}{ll}
0 & 5 \\
8 & 4
\end{array}\right]
\end{aligned}
$$
which is not true. Thus for $x=\frac{-1}{3}, y=7$, thus matrices are not equal.
(c) For $y=7, x=\frac{-2}{3}$
Thus for $y=7, x=\frac{-2}{3}$, the two matrices are not equal.
$$
\begin{aligned}
&\text { (d) } x=\frac{-1}{3}, y=\frac{-2}{3} \\
&{\left[\begin{array}{cc}
3 \times\left(\frac{-1}{3}\right)+7 & 5 \\
\frac{-2}{3}+1 & 2-3 \times\left(\frac{-1}{3}\right)
\end{array}\right]=\left[\begin{array}{cc}
0 & \frac{-2}{3}-2 \\
8 & 4
\end{array}\right]} \\
&\Rightarrow\left[\begin{array}{ll}
6 & 5 \\
\frac{1}{3} & 3
\end{array}\right]=\left[\begin{array}{cc}
0 & \frac{-8}{3} \\
8 & 4
\end{array}\right]
\end{aligned}
$$
which is not ture.
Hence for $x=\frac{-1}{3}, y=\frac{-2}{3}$, the two matrices are not equal.
Hence the option (b) not possible to find, is true.