Question

The perpendicular distance from the origin to the plane containing the two lines, $\frac{x+2}{3} = \frac{y-2}{5}=\frac{z+5}{7}$ and $\frac{x-1}{1} = \frac{y-4}{4} = \frac {z+4}{7}$, is $\frac{p}{\sqrt{q}}$, then $pq$ is

Answer 1

plane containing both lines
$ \therefore $ D.R. of plane $=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\ 3&5&7\\ 1&4&7\end{vmatrix}=7\hat{i}-14\hat{j}+7\hat{k} $
Now, equation of plane is,
$7 \left(x-1\right)-14 \left(y-4\right)+7 \left(z+4\right)=0 $
$\Rightarrow \Rightarrow x-1-2y+8+z+4=0$br> $\Rightarrow \Rightarrow x-2y+z+11=0$
Hence, distance from $\left(0, 0, 0\right)$ to the plane,
$=\frac{11}{\sqrt{1+4+1}}=\frac{11}{\sqrt{6}}$
$\because$ The given distance $=\frac{p}{\sqrt{q}}$
$\therefore \frac{p}{\sqrt{q}}=\frac{11}{\sqrt{6}} $
$\Rightarrow p=11, q=6$
Hence, $pq = 11\times6=66$