The lines $$\frac{x-2}{2} = \frac{y}{-2} = \frac{z-7}{16}$$ and $$\frac{x+3}{4} = \frac{y+2}{3} = \frac{z+2}{1}$$ intersect at the point P. If the distance of P from the line $$\frac{x+1}{2} = \frac{y-1}{3} = \frac{z-1}{1}$$ is $$l$$, then $$14l^2$$ is equal to _____.

Line 1: $$(2+2s, -2s, 7+16s)$$. Line 2: $$(-3+4t, -2+3t, -2+t)$$.

Setting equal: $$2s - 4t = -5$$ ... (i), $$2s + 3t = 2$$ ... (ii), $$16s - t = -9$$ ... (iii).

Subtracting (i) from (ii): $$7t = 7$$, so $$t = 1$$. From (ii): $$s = -1/2$$. Check (iii): $$-8 - 1 = -9$$ \checkmark

Point $$P = (1, 1, -1)$$.

Distance from P to line $$\frac{x+1}{2} = \frac{y-1}{3} = \frac{z-1}{1}$$:

Point $$A(-1, 1, 1)$$, direction $$\vec{d} = (2, 3, 1)$$, $$\vec{AP} = (2, 0, -2)$$.

$$\vec{AP} \times \vec{d} = (6, -6, 6)$$, $$|\vec{AP} \times \vec{d}|^2 = 108$$.

$$l^2 = 108/14$$. $$14l^2 = \boxed{108}$$.