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

Answer: 66

plane containing both lines
$\therefore$ D.R. of plane $=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 5& 7\\ 1& 4& 7\end{array}\right|=7\hat{i}-14\hat{j}+7\hat{k}$
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, $pq=11\times 6=66$