${\pi }_1$ is a plane passing through the point $(1,2,3)$ and perpendicular to the planes $x+2y+3z-6=0, x+2y+2z-5=0$. If $(-1,2,-3)$ is the foot of the perpendicular drawn from the point $(1,3,2)$ on to a plane ${\pi }_2,$ then the angle between the planes ${\pi }_1$ and ${\pi }_2$ is

We know that angle between two planes is the same as the angle between normals of these planes 

So normal vector of plane ${\mathrm{\pi }}_2$ is $\overrightarrow{n_2}=\left(-\hat{i}+2\hat{j}-3\hat{k}\right)-\left(\hat{i}+3\hat{j}+2\hat{k}\right)=-2\hat{i}-\hat{j}-5\hat{k}$

Now let direction ratios of normal of plane ${\mathrm{\pi }}_1$ are $a, b, c$

Given that ${\mathrm{\pi }}_1$ is perpendicular to planes 

$x+2y+3z-6=0 \& x+2y+2z-5=0$

So $a+2b+3c=0 \& a+2b+2c=0$

Now $\frac{a}{-2}=\frac{b}{1}=\frac{c}{0}$

So $\overrightarrow{n_1}=-2\hat{i}+\hat{j}$

Now angle between planes $\cos \theta =\frac{\overrightarrow{n_1}·\overrightarrow{n_2}}{\left|\overrightarrow{n_1}\right|\left|\overrightarrow{n_2}\right|}$

$=\frac{4-1}{\sqrt{5}\sqrt{30}}=\frac{3}{5\sqrt{6}}=\frac{\sqrt{6}}{10}$

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{\sqrt{6}}{10}\right)$