The line $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is parallel to the plane:

The plane $2x - y + 3z + 5 = 0$ is rotated through $90^{\circ}$ about its line of intersection with the plane $x + y + z = 1$. The equation of the plane in its new position is:

In $\mathbb{R}^3$,let $L$ be a straight line passing through the origin. Suppose that all the points on $L$ are at a constant distance from the two planes $P_1: x+2y-z+1=0$ and $P_2: 2x-y+z-1=0$. Let $M$ be the locus of the feet of the perpendiculars drawn from the points on $L$ to the plane $P_1$. Which of the following points lie$(s)$ on $M$?
$(A) \left(0, -\frac{5}{6}, -\frac{2}{3}\right)$
$(B) \left(-\frac{1}{6}, -\frac{1}{3}, \frac{1}{6}\right)$
$(C) \left(-\frac{5}{6}, 0, \frac{1}{6}\right)$
$(D) \left(-\frac{1}{3}, 0, \frac{2}{3}\right)$

The direction ratios of the line of intersection of the planes $x-y+z-5=0$ and $x-3y-6=0$ are:

The angle $\theta$ between the line $x - 1 = \frac{y - 2}{-1} = \frac{z - 3}{2}$ and the plane $\vec{r} \cdot (2\hat{i} + \hat{j} + \hat{k}) = 10$ is

Find the ratio in which the plane $x - 2y + 3z = 17$ divides the line segment joining the points $(-2, 4, 7)$ and $(3, -5, 8)$.