(a) Given, $\frac{42-13 x}{x^2+x-6}=\frac{\mathrm{A}}{l x+m}+\frac{\mathrm{B}}{p x+q}$
Take $\frac{42-13 x}{x^2+x-6}=\frac{42-13 x}{x^2+3 x-2 x-6}=\frac{42-13 x}{(x+3)(x-2)}$
$\begin{aligned}
& \frac{42-13 x}{(x+3)(x-2)}=\frac{\mathrm{A}}{(x+3)}+\frac{\mathrm{B}}{x-2} \\
& 42-13 x=\mathrm{A}(x-2)+\mathrm{B}(x+3)
\end{aligned}$
Put $x=2$,
$\begin{aligned}
& 42-26=5 \mathrm{~B} \\
& 5 \mathrm{~B}=16 \\
& \mathrm{~B}=\frac{16}{5}
\end{aligned}$
Now, put $x=-3$
$\begin{aligned}
& 42+39=-5 \mathrm{~A} \\
& 81=-5 \mathrm{~A} \\
& \mathrm{~A}=-\frac{81}{5}
\end{aligned}$
From (i),
$\frac{42-13 x}{(x+3)(x-2)}=\frac{-81}{5}\left(\frac{1}{x+3}\right)+\frac{16}{5(x-2)}$
Here, $l=1, m=3, \mathrm{p}=1, q=-2, \mathrm{~A}=\frac{-81}{5}, \mathrm{~B}=\frac{16}{5}$.
Take, $\frac{\mathrm{A} / \mathrm{p}}{\mathrm{B} m q}=\frac{\left(\frac{-81}{5}\right) \times 1 \times 1}{\left(\frac{16}{5}\right) \times 3 \times(-2)}=\frac{81}{16 \times 3 \times 2}=\frac{27}{32}$
Therefore, option (a) is correct.