Let the foot of perpendicular of the point $P\left(3, –2, –9\right)$ on the plane passing through the points $\left(–1, –2, –3\right), \left(9, 3, 4\right), \left(9, –2, 1\right)$ be $Q\left(\alpha , \beta , \gamma \right)$. Then the distance $Q$ from the origin is

Given that three points lie on a plane,

 

$\overrightarrow{AC}=10\hat{i}+4\hat{k}$

$\overrightarrow{AB}=10\hat{i}+5\hat{j}+7\hat{k}$

$\overrightarrow{AC}\times \overrightarrow{AB}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 10& 0& 4\\ 10& 5& 7\end{array}\right|$

$\Rightarrow \overrightarrow{AC}\times \overrightarrow{AB}=-20\hat{i}-30\hat{j}+50\hat{k}$

$\Rightarrow \overrightarrow{AC}\times \overrightarrow{AB}=2\hat{i}+3\hat{j}-5\hat{k}$

Equation of plane will be,

$\Rightarrow 2x+3y-5z=d$

And the plane passing through, $\left(–1, –2, –3\right)$

$\Rightarrow -2-6+15=d$

$\Rightarrow d=7$

$\therefore 2x+3y-5z=7$

Now finding the foot of perpendicular we get,

$\frac{x-3}{2}=\frac{y+2}{3}=\frac{z+9}{-5}=-\left(\frac{38}{38}\right)$

$x=1, y=-5, z=-4$

$Q\left(1, –5, –4\right)$

Distance from origin $=\sqrt{1+25+16}$$=\sqrt{42}$

Hence this is the required option.