If the line segment joining the points $P(2, 4, 1)$ and $Q(3, 8, 1)$ is divided by the plane $3x - ky - 6z = 0$ externally in the ratio $4:5$,then $k=$

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

Equation of the plane which passes through the point of intersection of lines $\frac{x - 1}{3} = \frac{y - 2}{1} = \frac{z - 3}{2}$ and $\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ and has the largest distance from the origin is

The image of the point $(-1, 3, 4)$ in the plane $x - 2y = 0$ is

Let the image of the point $P(2, -1, 3)$ in the plane $x + 2y - z = 0$ be $Q$. Then the distance of the plane $3x + 2y + z + 29 = 0$ from the point $Q$ is $.........$.

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