Let $$P$$ be the plane containing the straight line $$\dfrac{x - 3}{9} = \dfrac{y + 4}{-1} = \dfrac{z - 7}{-5}$$ and perpendicular to the plane containing the straight lines $$\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z}{5}$$ and $$\dfrac{x}{3} = \dfrac{y}{7} = \dfrac{z}{8}$$. If $$d$$ is the distance of $$P$$ from the point $$(2, -5, 11)$$, then $$d^2$$ is equal to

We need to find the distance squared from point $$(2, -5, 11)$$ to plane $$P$$.

Line 1: $$\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z}{5}$$ has direction $$\vec{d_1} = (2, 3, 5)$$.

Line 2: $$\dfrac{x}{3} = \dfrac{y}{7} = \dfrac{z}{8}$$ has direction $$\vec{d_2} = (3, 7, 8)$$.

Normal to their plane:

$$\vec{n_1} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 3 & 5 \\ 3 & 7 & 8 \end{vmatrix} = (24 - 35)\vec{i} - (16 - 15)\vec{j} + (14 - 9)\vec{k} = (-11, -1, 5)$$

Plane $$P$$ contains the line $$\dfrac{x-3}{9} = \dfrac{y+4}{-1} = \dfrac{z-7}{-5}$$ (direction $$\vec{d_3} = (9, -1, -5)$$) and is perpendicular to the plane with normal $$\vec{n_1} = (-11, -1, 5)$$.

The normal to $$P$$ is perpendicular to both $$\vec{d_3}$$ and $$\vec{n_1}$$:

$$\vec{n} = \vec{d_3} \times \vec{n_1} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 9 & -1 & -5 \\ -11 & -1 & 5 \end{vmatrix}$$ $$= (-5 - 5)\vec{i} - (45 - 55)\vec{j} + (-9 - 11)\vec{k} = (-10, 10, -20)$$

Simplify: $$\vec{n} = (-1, 1, -2)$$ (dividing by 10).

$$P$$ passes through $$(3, -4, 7)$$ with normal $$(-1, 1, -2)$$:

$$-1(x - 3) + 1(y + 4) - 2(z - 7) = 0$$ $$-x + 3 + y + 4 - 2z + 14 = 0$$ $$-x + y - 2z + 21 = 0$$

Or equivalently: $$x - y + 2z = 21$$.

$$d = \dfrac{|2 - (-5) + 2(11) - 21|}{\sqrt{1 + 1 + 4}} = \dfrac{|2 + 5 + 22 - 21|}{\sqrt{6}} = \dfrac{|8|}{\sqrt{6}} = \dfrac{8}{\sqrt{6}}$$ $$d^2 = \dfrac{64}{6} = \dfrac{32}{3}$$

The answer is Option C: $$\dfrac{32}{3}$$.