A line with direction ratio $$2, 1, 2$$ meets the lines $$x = y + 2 = z$$ and $$x + 2 = 2y = 2z$$ respectively at the point $$P$$ and $$Q$$. If the length of the perpendicular from the point $$(1, 2, 12)$$ to the line $$PQ$$ is $$l$$, then $$l^2$$ is _______

We need to find $$l^2$$, where $$l$$ is the perpendicular distance from $$(1, 2, 12)$$ to line $$PQ$$.

Parametrise the given lines.

Line 1: $$x = y + 2 = z$$. Let $$x = t$$, then $$y = t-2$$, $$z = t$$. Direction: $$(1, 1, 1)$$. Point: $$(t, t-2, t)$$.

Line 2: $$x + 2 = 2y = 2z$$. Let $$2y = s$$, then $$y = s/2$$, $$z = s/2$$, $$x = s-2$$. Direction: $$(2, 1, 1)$$. Point: $$(s-2, s/2, s/2)$$.

Find P and Q.

The line through P and Q has direction ratios $$2:1:2$$.

P is on Line 1: $$P = (t, t-2, t)$$. Q is on Line 2: $$Q = (s-2, s/2, s/2)$$.

Direction $$PQ$$: $$(s-2-t, s/2-t+2, s/2-t)$$ must be proportional to $$(2, 1, 2)$$.

$$\frac{s-2-t}{2} = \frac{s/2-t+2}{1} = \frac{s/2-t}{2} = k$$

From equations 1 and 3: $$s-2-t = s/2-t$$, so $$s/2 = 2$$, $$s = 4$$.

From equation 3: $$k = (2-t)/2$$.

From equation 2: $$2-t+2 = k = (2-t)/2$$, so $$4-t = (2-t)/2$$, $$8-2t = 2-t$$, $$t = 6$$.

So $$P = (6, 4, 6)$$ and $$Q = (2, 2, 2)$$.

Find the perpendicular distance from $$(1, 2, 12)$$ to line PQ.

Direction of PQ: $$(6-2, 4-2, 6-2) = (4, 2, 4)$$ or simplified $$(2, 1, 2)$$.

$$\vec{AQ} = (1-2, 2-2, 12-2) = (-1, 0, 10)$$ where $$A = (1, 2, 12)$$.

$$\vec{AQ} \times \vec{d} = (-1, 0, 10) \times (2, 1, 2)$$

$$= (0 \cdot 2 - 10 \cdot 1, 10 \cdot 2 - (-1) \cdot 2, (-1) \cdot 1 - 0 \cdot 2) = (-10, 22, -1)$$

$$|\vec{AQ} \times \vec{d}| = \sqrt{100 + 484 + 1} = \sqrt{585}$$

$$|\vec{d}| = \sqrt{4+1+4} = 3$$

$$l = \frac{\sqrt{585}}{3}$$

$$l^2 = \frac{585}{9} = 65$$

The answer is 65.