The angle between the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and the plane $3x + 2y - 3z = 4$ is ......... $^o$

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:

The image of the point $(5, 2, 6)$ with respect to the plane $x + y + z = 9$ is

If $\theta$ is the angle between the line $\frac{x + 1}{3} = \frac{y - 2}{2} = \frac{z - 2}{4}$ and the plane $2x + y - 3z + 4 = 0$,then $64 \csc^2 \theta$ is equal to

The lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{-k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar if

Let $P_1$ and $P_2$ be two planes given by $P_1: 10x + 15y + 12z - 60 = 0$ and $P_2: -2x + 5y + 4z - 20 = 0$. Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on $P_1$ and $P_2$?
$(A) \frac{x-1}{0} = \frac{y-1}{0} = \frac{z-1}{5}$
$(B) \frac{x-6}{-5} = \frac{y}{2} = \frac{z}{3}$
$(C) \frac{x}{-2} = \frac{y-4}{5} = \frac{z}{4}$
$(D) \frac{x}{1} = \frac{y-4}{-2} = \frac{z}{3}$