The equation of the plane containing the line $2x - 5y + z = 3; x + y + 4z = 5$ and parallel to the plane $x + 3y + 6z = 1$ is:

Let $P(2,1,5)$ be a point in space and $Q$ be a point on the line $\vec{r}=(\hat{i}-\hat{j}+2\hat{k})+\mu(-3\hat{i}+\hat{j}+5\hat{k})$. Then the value of $\mu$ for which the vector $\vec{PQ}$ is parallel to the plane $3x-y+4z=1$ is

Find the distance of the point $-\hat{i} - 5\hat{j} - 10\hat{k}$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 5$.

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 $M$?
$(A) \left(0, -\frac{5}{6}, -\frac{2}{3}\right)$
$(B) \left(-\frac{1}{6}, -\frac{1}{3}, \frac{1}{6}\right)$
$(C) \left(-\frac{5}{6}, 0, \frac{1}{6}\right)$
$(D) \left(-\frac{1}{3}, 0, \frac{2}{3}\right)$

What is the point of intersection of the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z + 2}{3}$ and the plane $2x + 3y + z = 0$?

The length of the projection of the line segment joining the points $(5, -1, 4)$ and $(4, -1, 3)$ on the plane $x + y + z = 7$ is: