Mathematics - Vector Question with Solution | TestHub

Let $l, m, n$ be the direction cosines of the line normal to the plane, then

$l-2m+3n=0$ and $2l-m-n=0$

$\Rightarrow \frac{l}{2+3}=\frac{m}{6+1}=\frac{n}{-1+4}=\lambda$



$\therefore$ The equation of the plane is

$5x+7y+3y+d=0$

$\because$it passes through$\left(1,-1,-1\right)$

$\Rightarrow 5-7-3+d=0$

$\Rightarrow d=5$

Hence, the equation of the plane is $5x+7y+3y+5=0$

Now, $\left|PQ\right|=\frac{\left|5+21-21+5\right|}{\sqrt{{\left(5\right)}^2+{\left(7\right)}^2+{\left(3\right)}^2}}$

$\Rightarrow \left|PQ\right|=\frac{10}{\sqrt{25+49+9}}$

$\Rightarrow \left|PQ\right|=\frac{10}{\sqrt{83}}$