Let the points on the plane $P$ be equidistant from the points $\left(-4,2,1\right)$ and $\left(2,-2,3\right)$. Then the acute angle between the plane $P$ and the plane $2x+y+3z=1$ is

Plotting the diagram of the given information we get,

Normal vector of plane will be $=\overrightarrow{AB}=\left(\overrightarrow{OB}-\overrightarrow{OA}\right)$

$=\left(6\hat{i}-4\hat{j}+2\hat{k}\right)$ $=2\left(3\hat{i}-2\hat{j}+\hat{k}\right)$

Now equation of plane is given by,

$P\equiv 3\left(x+1\right)-2\left(y\right)+1\left(z-2\right)=0$

$P\equiv 3x-2y+z+1=0$

Assuming other plane to be $P^{\text{'}}\equiv 2x+y+3z-1=0$

Now the angle between them is given by, $\cos \theta =\left|\frac{{\hat{n}}_1·{\hat{n}}_2}{\left|n_1\right|\left|n_2\right|}\right|$

$\theta ={\cos }^{-1}\left(\frac{6-2+3}{\sqrt{14}\times \sqrt{14}}\right)$

$\theta ={\cos }^{-1}\left(\frac{7}{14}\right)=={\cos }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{3}$