Find the equation of the line perpendicular to the plane $2x + 4y - 5z = 10$ passing through the origin.

The value of $k$ such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$ lies in the plane $2x-4y+z=7$ is:

Let $S$ be the reflection of a point $Q$ with respect to the plane given by $\vec{r} = -(t+p) \hat{i} + \hat{j} + (1+p) \hat{k}$,where $t, p$ are real parameters and $\hat{i}, \hat{j}, \hat{k}$ are the unit vectors along the three positive coordinate axes. If the position vectors of $Q$ and $S$ are $10 \hat{i} + 15 \hat{j} + 20 \hat{k}$ and $\alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$ respectively,then which of the following is/are $TRUE$?
$(A)$ $3(\alpha+\beta) = -101$
$(B)$ $3(\beta+\gamma) = -71$
$(C)$ $3(\gamma+\alpha) = -86$
$(D)$ $3(\alpha+\beta+\gamma) = -121$

Let $\vec{r} \cdot (3\hat{i} - 2\hat{j} + 7\hat{k}) = 32$ be the equation of a plane. If the line with direction ratios $(5, b, 3)$ is parallel to the plane, then the value of $b$ is:

The line $\frac{x - 1}{2} = -(y + 1) = \frac{z}{3}$ and the plane $3x + 2y - z = 5$ intersect at a point. The coordinates of the point are:

If the distance between the plane $Ax-2y+z=d$ and the plane containing the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is $\sqrt{6}$ units,then $|d|$ is