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

The equation of the plane passing through the line of intersection of the planes $\overline{r} \cdot(2 \hat{i}-3 \hat{j}+4 \hat{k})=1$ and $\overline{r} \cdot(\hat{i}-\hat{j})+4=0$,and perpendicular to the plane $\overline{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})+8=0$,is given by $\overline{r} \cdot(-5 \hat{i}+2 \hat{j}+12 \hat{k})=\mu$. Then the value of $\mu$ is:

The equation of the plane containing the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and the point $(0,7,-7)$ is

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

The equation of the plane containing the line of intersection of the planes $2x - y = 0$ and $y - 3z = 0$ and perpendicular to the plane $4x + 5y - 3z - 8 = 0$ is

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