Explanation:

(C): Let image of point \(\mathrm{P}(-1,2)\) is \(\mathrm{P}^{\prime}(\mathrm{a}, \mathrm{b})\).
\(\therefore \mathrm{PM}=\mathrm{MP}^{\prime} \& \mathrm{PP}^{\prime} \perp \mathrm{AB}\)
Given, \(\mathrm{P}^{\prime}(\mathrm{a}, \mathrm{b})\)
\(\therefore \mathrm{M}=\left(\frac{\mathrm{a}-1}{2}, \frac{\mathrm{b}+2}{2}\right)\)
It lies on the line \(x-2 y+3=0\)
\(\therefore\left(\frac{\mathrm{a}-1}{2}\right)-2\left(\frac{\mathrm{b}+2}{2}\right)+3=0\)
\(a-1-2 b-4+6=0\)
\(\mathrm{a}-2 \mathrm{~b}+1=0\)
slope of \(\mathrm{PP}^{\prime}=\frac{\mathrm{b}-2}{\mathrm{a}+1}\)
\(\mathrm{PP}^{\prime} \perp \mathrm{AB}\)
\(\frac{\mathrm{b}-2}{\mathrm{a}+1} \times \frac{1}{2}=-1\)
\(\mathrm{b}-2=-2 \mathrm{a}-2\)
\(2 \mathrm{a}+\mathrm{b}=0\)
From equation (i) \& (ii) we get -
\(b=2 / 5\)
\(a=-1 / 5\)
\(\therefore\) Image of \(\mathrm{P}(-1,2)\) is \(\mathrm{P}^{\prime}\left(\frac{-1}{5}, \frac{2}{5}\right)\)
length of perpendicular from \(\mathrm{P}^{\prime}\) on to the line \(2 \mathrm{x}+\mathrm{y}-7=0\) is
\(\mathrm{d}=\frac{\left|2 \times \frac{-1}{5}+\frac{2}{5}-7\right|}{\sqrt{2^{2}+1}}=\frac{\left|\frac{-2}{5}+\frac{2}{5}-7\right|}{\sqrt{5}}\)
\(\mathrm{d}=\frac{7}{\sqrt{5}}\)