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 $\left|\overrightarrow{CD}\right|^2$ is equal to:

Let the values of $\lambda$ for which the shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$
and $\frac{x-\lambda}{3} = \frac{y-4}{4} = \frac{z-5}{5}$ is $\frac{1}{\sqrt{6}}$ be $\lambda_1$ and $\lambda_2$. Then the radius of the circle passing through the
points $(0, 0), (\lambda_1, \lambda_2)$ and $(\lambda_2, \lambda_1)$ is

If the equation of the line passing through the point $ \left( 0, -\frac{1}{2}, 0 \right) $ and perpendicular to the lines $ \vec{r} = \lambda \left( \hat{i} + a\hat{j} + b\hat{k} \right) $ and $ \vec{r} = \left( \hat{i} - \hat{j} - 6\hat{k} \right) + \mu \left( -b \hat{i} + a\hat{j} + 5\hat{k} \right) $ is $ \frac{x-1}{-2} = \frac{y+4}{d} = \frac{z-c}{-4} $, then $ a+b+c+d $ is equal to :

Consider the lines L₁: x - 1 = y - 2 = z and L₂: x - 2 = y = z - 1. Let the feet of the perpendiculars from the point P(5, 1, -3) on the lines L₁ and L₂ be Q and R respectively. If the area of the triangle PQR is A, then 4A² is equal to :