The image of the point $(3, 2, 1)$ in the plane $2x - y + 3z = 7$ is

$A$ plane is making intercepts $2, 3, 4$ on $X, Y$ and $Z$-axes respectively. Another plane is passing through the point $(-1, 6, 2)$ and is perpendicular to the line joining the points $(1, 2, 3)$ and $(-2, 3, 4)$. Then the angle between the two planes is

The direction cosines of the line formed by the intersection of the planes $x - y + 2z = 5$ and $3x + y + z = 6$ are:

Let $P$ be a point in the first octant,whose image $Q$ in the plane $x+y=3$ (that is,the line segment $PQ$ is perpendicular to the plane $x+y=3$ and the mid-point of $PQ$ lies in the plane $x+y=3$) lies on the $z$-axis. Let the distance of $P$ from the $x$-axis be $5$. If $R$ is the image of $P$ in the $xy$-plane,then the length of $PR$ is.

The distance of the point $P(3,4,4)$ from the point of intersection of the line joining the points $Q(3,-4,-5)$ and $R(2,-3,1)$ and the plane $2x+y+z=7$ is equal to $.....$

Find the angle between the line $\vec{r} = (\hat{i} + 2\hat{j} - \hat{k}) + \lambda(\hat{i} + \hat{j} - \hat{k})$ and the plane $\vec{r} \cdot (-2\hat{i} + \hat{j} - \hat{k}) = 0$.