If $P^{\text{'}}(a, b)$ is the image of the point $P(-1, 2)$ with respect to the line $x-2y+3=0,$ then the length of the perpendicular from $P^{\text{'}}$ on to the line $2x+y-7=0$ is

The image $\left(h,k\right)$ of a point $P\left(x_1,y_1\right)$ about the line $ax+by+c=0$ is given  by

$\frac{h-x_1}{a}=\frac{k-y_1}{b}=-2\frac{\left(a{x}_1+b{y}_1+c\right)}{a^2+b^2}$

Now putting all values from the given data in this formula, we get

$\frac{a+1}{1}=\frac{b-2}{-2}=\frac{-2\left(-1-4+3\right)}{1+4}$

$a=\frac{-1}{5},b=\frac{2}{5}$

Length of perpendicular from $\left(\frac{-1}{5},\frac{2}{5}\right)$ on the line $2x+y-7=0$ is given by

$\left|\frac{2\times {\displaystyle \frac{-1}{5}}+{\displaystyle \frac{2}{5}}-7}{\sqrt{1+4}}\right|=\left|\frac{-7}{\sqrt{5}}\right|=\frac{7}{\sqrt{5}}$