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 :

Let the line L pass through $(1,1,1)$ and intersect the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z}{1}$. Then, which of the following points lies on the line $L$ ?