Thank you for reporting, we will resolve it shortly
Q.
Let the points on the plane $P$ be equidistant from the points $(-4,2,1)$ and $(2,-2,3)$. Then the acute angle between the plane $P$ and the plane $2 x + y +$ $3 z =1$ is
Normal vector $=\overrightarrow{ AB }=(\overrightarrow{ OB }-\overrightarrow{ OA })$
$=(6 \hat{ i }-4 \hat{ j }+2 \hat{ k })$
or $2(3 \hat{ i }-2 \hat{ j }+\hat{ k })$
$P \equiv 3( x +1)-2( y )+1( z -2)=0$
$P \equiv 3 x -2 y + z +1=0$
$P ^{\prime} \equiv 2 x + y +3 z -1=0$
angle between $P \& P ^{\prime}=\left|\frac{\hat{n}_{1} \cdot \hat{n}_{2}}{\left| n _{1}\right|\left| n _{2}\right|}\right|=\cos \theta$
$\theta=\cos ^{-1}\left(\frac{6-2+3}{\sqrt{14} \times \sqrt{14}}\right)$
$\theta=\cos ^{-1}\left(\frac{7}{14}\right)= \cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$
