Let the line $L_1$ be parallel to the vector $-3\hat{i} + 2\hat{j} + 4\hat{k}$ and pass through the point $(2, 6, 7)$, and the line $L_2$ be parallel to the vector $2\hat{i} + \hat{j} + 3\hat{k}$ and pass through the point $(4, 3, 5)$. If the line $L_3$ is parallel to the vector $-3\hat{i} + 5\hat{j} + 16\hat{k}$ and intersects the lines $L_1$ and $L_2$ at the points $C$ and $D$, respectively, then $|CD|^2$ is equal to:

$L_1 : \frac{x-2}{-3} = \frac{y-6}{2} = \frac{z-7}{4}$

Point $C$ on $L_1$: $(-3\lambda_1 + 2,; 2\lambda_1 + 6,; 4\lambda_1 + 7)$

$L_2 : \frac{x-4}{2} = \frac{y-3}{1} = \frac{z-5}{3}$

Point $D$ on $L_2$: $(2\lambda_2 + 4,; \lambda_2 + 3,; 3\lambda_2 + 5)$

D.R’s of line $L_3$:

$\frac{2\lambda_1 + 3\lambda_1 + 2}{-3} = \frac{\lambda_2 - 2\lambda_1 - 3}{5} = \frac{3\lambda_2 - 4\lambda_1 - 2}{16}$

$\lambda_1 = -3,; \lambda_2 = 2$

$C(11, 0, -5)$

$D(8, 5, 11)$

$|CD|^2 = 3^2 + 5^2 + 16^2 = 290$