A plane $$P$$ contains the line of intersection of the plane $$\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 6$$ and $$\vec{r} \cdot (2\hat{i} + 3\hat{j} + 4\hat{k}) = -5$$. If $$P$$ passes through the point (0, 2, -2), then the square of distance of the point (12, 12, 18) from the plane P is

Plane P passes through the intersection of $$x+y+z = 6$$ and $$2x+3y+4z = -5$$, and through $$(0, 2, -2)$$.

Write the family of planes.

$$(x+y+z-6) + \lambda(2x+3y+4z+5) = 0$$

$$(1+2\lambda)x + (1+3\lambda)y + (1+4\lambda)z + (-6+5\lambda) = 0$$

Use the point (0, 2, -2).

$$0 + (1+3\lambda)(2) + (1+4\lambda)(-2) + (-6+5\lambda) = 0$$

$$2 + 6\lambda - 2 - 8\lambda - 6 + 5\lambda = 0$$

$$3\lambda - 6 = 0$$

$$\lambda = 2$$

Find the equation of plane P.

$$(1+4)x + (1+6)y + (1+8)z + (-6+10) = 0$$

$$5x + 7y + 9z + 4 = 0$$

Find distance from (12, 12, 18).

$$d = \dfrac{|5(12) + 7(12) + 9(18) + 4|}{\sqrt{25 + 49 + 81}} = \dfrac{|60 + 84 + 162 + 4|}{\sqrt{155}} = \dfrac{310}{\sqrt{155}}$$

$$d^2 = \dfrac{310^2}{155} = \dfrac{96100}{155} = 620$$

The answer is $$\boxed{620}$$ (Option 1). This matches the stored answer.