Let P he the plane passing through the point (1, 2, 3) and perpendicular to the line$\frac{x-2}{1}=\frac{y-3}{-1}=\frac{z-4}{2}$. Then the distance of the (3, -2, -3) from the P is:

$\sqrt{6}$

Line: $\frac{x-2}{1}=\frac{y-3}{-1}=\frac{z-4}{2}$

Direction ratios of the line $=(1,-1,2)$.

The plane is perpendicular to this line, so its normal vector is $(1,-1,2)$.

Plane passes through $(1,2,3)$.

Equation of plane:

$1(x-1)-1(y-2)+2(z-3)=0$

$x-y+2z-5=0$

Distance of point $(3,-2,-3)$ from plane:

$=\frac{|3-(-2)+2(-3)-5|}{\sqrt{1^2+(-1)^2+2^2}}$

$=\frac{|3+2-6-5|}{\sqrt{6}}$

$=\frac{6}{\sqrt{6}}$

$=\sqrt{6}$

The required distance is $\sqrt{6}$.