Explanation

(d)
Let $B(x_{1},\ y_{1})$ and $C(x_{2},y_{2})$ are the vertices of a triangle
$P\left( \frac{x_{1} + 1}{2},\frac{y_{1} - 2}{2} \right)\ lies\ on\ the\ line\ x - y + 5 = 0$
$\therefore\frac{x_{1} + 1}{2} - \frac{y_{1} - 2}{2} = - 5$
$\Rightarrow x_{1} - y_{1} = - 13$ …(i)
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Also, $PN\bot AB$
$\therefore\frac{y_{1} + 2}{x_{1} - 1} = - 1$
$\Rightarrow y_{1} + 2 = - x_{1} + 1$
$\Rightarrow x_{1} + y_{1} = - 1$ …(ii)
On solving Eqs. (i) and (ii), we get
$x_{1} = - 7\ and\ y_{1} = 6$
$\therefore$ Coordinates of $B$ are $( - 7,\ 6)$
Similarly, the coordinates of $C$ are $\left( \frac{11}{5},\frac{2}{5} \right)$
$\therefore$ Equation of $BC$ is
$(y - 6) = \frac{\frac{2}{5} - 6}{\frac{11}{5} + 7}(x + 7)$
$\Rightarrow 14x + 23y - 40 = 0$