Question:
If the equation of a plane $\mathrm{P}$, passing through the intesection of the planes, $x+4 y-z+7=0$ and $3 x+y+5 z=8$ is $a x+b y+6 z=15$ for some $a, b \in R$, then the distance of the point $(3,2,$, -1) from the plane $\mathrm{P}$ is
Solution:
$\mathrm{D}_{1}=\left|\begin{array}{ccc}-7 & 4 & -1 \\ 8 & 1 & 5 \\ 15 & \mathrm{~h} & 6\end{array}\right|=0 \Rightarrow \mathrm{b}=-3$
$\mathrm{D}=\left|\begin{array}{ccc}1 & 4 & -1 \\ 3 & 1 & 5 \\ \mathrm{a} & \mathrm{b} & 6\end{array}\right|=0 \Rightarrow 21 \mathrm{a}-8 \mathrm{~b}-66=0 \ldots .$ (1)
$P: 2 x-3 y+6 z=15$
so required distance $=\frac{21}{7}=3$