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:

$\frac{1}{\sqrt{6}}$

The required plane is perpendicular to the line  (x-2)/1= (y-3)/-1 = (z-7)/2 and passes through a point (1,2,3)

As we know that the equation of the plane perpendicular to a line with direction ratios (a,b,c) and passing through the point (x_1,   y_1,  z₁) 

a (x- x₁) + b (y-y₁) + c(z-z₁₎ = 0

Here a = 1 , b= -1, c= 2 and X₁ = 1, y₁ = 2,  z₁= 3

So, the equation of the plane 

1(x- 1)  - 1(y- 2) + 2(z - 3) =0

x- y + 2z -5 =0

The distance between the plane x- y + 2z -5 =0  and the point (3, -2 ,-3) 

  = {(1×3)  + (-1×-2) +(2 ×-3)}/√(6)

  = 1/√(6)