If the line $\frac{x - 3}{1} = \frac{y + 2}{-1} = \frac{z + \lambda}{-2}$ lies in the plane $2x - 4y + 3z = 2$,then the shortest distance between this line and the line $\frac{x - 1}{12} = \frac{y}{9} = \frac{z}{4}$ is

Let $P(3, 2, 6)$ be a point in space and $Q$ be a point on the line $\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(-3\hat{i} + \hat{j} + 5\hat{k})$. Then the value of $\mu$ for which the vector $\vec{PQ}$ is parallel to the plane $x - 4y + 3z = 1$ is

Find the equation of the plane which contains the line of intersection of the planes $\vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})-4=0$ and $\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})+5=0$ and which is perpendicular to the plane $\vec{r} \cdot(5 \hat{i}+3 \hat{j}-6 \hat{k})+8=0$.

What is the equation of the line passing through $(1, 2, 3)$ and perpendicular to the plane $3x + 4y - 5z = 6$?

$A$ line with positive direction cosines passes through the point $P(2,-1,2)$ and makes equal angles with the coordinate axes. The line meets the plane $2x+y+z=9$ at point $Q$. The length of the line segment $PQ$ is equal to:

Let the image of the point $P(1, 2, 3)$ in the plane $2x - y + z = 9$ be $Q$. If the coordinates of the point $R$ are $(6, 10, 7)$,then the square of the area of the triangle $PQR$ is $.....$.