Let a plane $P$ pass through the point $\left(3,7,-7\right)$ and contain the line, $\frac{x-2}{-3}=\frac{y-3}{2}=\frac{z+2}{1}.$ If distance of the plane $P$ from the origin is $d,$ then $d^2$ is equal to

The required plane passes through the point $\left(3,7,-7\right).$

Assume, the equation of the plane is $a\left(x-3\right)+b\left(y-7\right)+c\left(z+7\right)=0$

The line $\frac{x-2}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ lies on the required plane.

So, we can say that 

$a\left(2-3\right)+b\left(3-7\right)+c\left(-2+7\right)=0$

$\Rightarrow -a-4b+5c=0 ...\left(1\right)$

And $-3\times a+2\times b+1\times c=0 ...\left(2\right)$

Use cross-multiplication to solve the equations $\left(1\right) \& \left(2\right)$

$\frac{a}{\left(-4\right)\times 1-2\times 5}=\frac{b}{5\times \left(-3\right)+1\times 1}=\frac{c}{\left(-1\right)\times 2-\left(-4\right)\times \left(-3\right)}$

$\Rightarrow \frac{a}{-14}=\frac{b}{-14}=\frac{c}{-14}=k$  (Assume)

So, $a=k, b=k, c=k$

Put the values of $a,b,c$ in the required equation of plane we get $\left(x-3\right)+\left(y-7\right)+\left(z+7\right)=0$

$\Rightarrow x+y+z=3$

So, the distance of the plane from the origin is $\frac{\left|3\right|}{\sqrt{1^2+1^2+1^2}}=\sqrt{3}$

Hence, $d^2={\left(\sqrt{3}\right)}^2=3$