If two distinct points $Q$ and $R$ lie on the line of intersection of the planes $-x + 2y - z = 0$ and $3x - 5y + 2z = 0$,and $PQ = PR = \sqrt{18}$,where the point $P$ is $(1, -2, 3)$,then the area of the triangle $PQR$ is equal to

The angle between the line $\frac{x - 2}{a} = \frac{y - 2}{b} = \frac{z - 2}{c}$ and the plane $ax + by + cz + 6 = 0$ is ......... $^o$

For what value of $c$ does the line $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{-1}$ intersect the curve $xy = c^2, z = 0$?

If the line passing through the points $(5, 1, a)$ and $(3, b, 1)$ intersects the plane at the point $(0, \frac{17}{2}, -\frac{13}{2})$,then find the values of $a$ and $b$.

$A$ line $L$ is passing through the point $A$ whose position vector is $\hat{i}+2 \hat{j}-3 \hat{k}$ and is parallel to the vector $2 \hat{i}+\hat{j}+2 \hat{k}$. $A$ plane $\pi$ is passing through the points $\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-\hat{j}-\hat{k}$ and is parallel to the vector $\hat{i}-2 \hat{j}$. Then the point where this plane $\pi$ meets the line $L$ is

The equation of the plane in which the lines $\frac{x - 5}{4} = \frac{y - 7}{4} = \frac{z + 3}{-5}$ and $\frac{x - 8}{7} = \frac{y - 4}{1} = \frac{z - 5}{3}$ lie,is