This can also be found from the cross product $\overrightarrow{\mathrm{AB}}\times \overrightarrow{\mathrm{AC}}=(-1,0,1)\times (-1,6,3)$.

Suppose the plane $P_2$ has equation $\mathrm{lx}+\mathrm{my}+\mathrm{nz}=1$, then for the planes $P_1$ and $P_2$ to be perpendicular the point $V$ must satisfy this equation. The normal vector to $P_2$, i.e. $(l,m,n)$ must be perpendicular to $(3,-1,3)$ so $3l-m+3n=0$. As $A$ and $B$ are on $P_2$ we have:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{2l+m}& =1\\ \multicolumn{1}{c}{l+m+n}& =1\\ \multicolumn{1}{c}{3l-m+3n}& =0\end{array}}$

Hence $l=1/8$, $m=3/4$ and $n=1/8$ so the equation of $P_2$ is $x+6y+z=8$. The coordinates of $V$ must satisfy this equation giving an infinite set of points on the plane $P_2$.