The angle between the line $r = (2i - j + k) + \lambda (-i + j + k)$ and the plane $r \cdot (3i + 2j - k) = 4$ is:

The equation of the plane passing through the point $(2,-1,-3)$ and parallel to the lines $\frac{x-1}{3}=\frac{y+2}{2}=\frac{z}{-4}$ and $\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}$ is

Let the image of the point $P(1, 6, a)$ in the line $L: \frac{x-1}{2} = \frac{y-2}{b} = \frac{z-a+1}{1}, b>0$, be $Q(\frac{a}{3}, 0, a+c)$. If $S(\alpha, \beta, \gamma), \alpha > 0$, is the point on $L$ such that the distance of $S$ from the foot of perpendicular $F$ from the point $P$ on $L$ is $2\sqrt{14}$, then $\alpha + \beta + \gamma$ is equal to:

If for $a > 0,$ the feet of perpendiculars from the points $A(a, -2a, 3)$ and $B(0, 4, 5)$ on the plane $lx + my + nz = 0$ are points $C(0, -a, -1)$ and $D$ respectively,then the length of line segment $CD$ is equal to

The line of intersection of the planes $r \cdot (i - 3j + k) = 1$ and $r \cdot (2i + 5j - 3k) = 2$ is parallel to which vector?

Find the equation of the plane passing through the line of intersection of the planes $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=1$ and $\vec{r} \cdot(2 \hat{i}+3 \hat{j}-\hat{k})+4=0$ and parallel to the $x$-axis.