Let the line $$L_{1}$$ be parallel to the vector $$-3\widehat{i} +2\widehat{j} + 4\widehat{k}$$ and pass through the point (2, 6, 7), and the line $$L_{2}$$ be parallel to the vector $$2\widehat{i} +\widehat{j} + 3\widehat{k}$$ and pass through the point (4, 3, 5). If the line $$L_{3}$$ is parallel to the vector $$-3\widehat{i} +5\widehat{j} + 16\widehat{k}$$ and intersects the lines $$L_{1}$$ and $$L_{2}$$ at the points C and D, respectively, then $$|\overrightarrow{CD}|^2$$ is equal to :

$$L_1$$: passes through $$(2,6,7)$$ with direction $$(-3, 2, 4)$$.

$$L_2$$: passes through $$(4,3,5)$$ with direction $$(2, 1, 3)$$.

$$L_3$$: direction $$(-3, 5, 16)$$, intersects $$L_1$$ at $$C$$ and $$L_2$$ at $$D$$.

Parametrize the lines.

$$C = (2-3s, 6+2s, 7+4s)$$ on $$L_1$$.

$$D = (4+2t, 3+t, 5+3t)$$ on $$L_2$$.

$$\vec{CD}$$ must be parallel to $$(-3, 5, 16)$$.

$$\vec{CD} = (4+2t-2+3s, 3+t-6-2s, 5+3t-7-4s) = (2+2t+3s, -3+t-2s, -2+3t-4s)$$

For proportionality with $$(-3, 5, 16)$$:

$$\frac{2+2t+3s}{-3} = \frac{-3+t-2s}{5} = \frac{-2+3t-4s}{16}$$

Let this ratio be $$k$$. Then:

$$2+2t+3s = -3k$$ ... (i)

$$-3+t-2s = 5k$$ ... (ii)

$$-2+3t-4s = 16k$$ ... (iii)

From (ii): $$k = \frac{-3+t-2s}{5}$$

From (i): $$2+2t+3s = -3 \cdot \frac{-3+t-2s}{5} = \frac{9-3t+6s}{5}$$

$$10+10t+15s = 9-3t+6s$$

$$13t + 9s = -1$$ ... (iv)

From (iii): $$-2+3t-4s = 16 \cdot \frac{-3+t-2s}{5} = \frac{-48+16t-32s}{5}$$

$$-10+15t-20s = -48+16t-32s$$

$$-t + 12s = -38$$

$$t - 12s = 38$$ ... (v)

From (v): $$t = 38 + 12s$$. Substituting in (iv):

$$13(38+12s) + 9s = -1$$

$$494 + 156s + 9s = -1$$

$$165s = -495$$

$$s = -3$$

$$t = 38 + 12(-3) = 2$$

Find C and D.

$$C = (2+9, 6-6, 7-12) = (11, 0, -5)$$

$$D = (4+4, 3+2, 5+6) = (8, 5, 11)$$

$$|\vec{CD}|^2 = (8-11)^2 + (5-0)^2 + (11+5)^2 = 9 + 25 + 256 = 290$$

The answer is Option 1: 290.