If the line $\frac{x+1}{2}=\frac{y-m}{3}=\frac{z-4}{6}$ lies in the plane $3x-14y+6z+49=0$,then the value of $m$ 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 passing through the intersection of the planes $ax + by + cz + d = 0$ and $a'x + b'y + c'z + d' = 0$ and parallel to the line $y = 0, z = 0$ is:

The sine of the angle between the straight line $\frac{x-2}{2}=\frac{y-3}{4}=\frac{4-z}{2}$ and the plane $2x-2y+z=5$ is

The equation of the plane passing through the intersection of the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-5}{-3}$ and $\frac{x+5}{3}=\frac{y-4}{-1}=\frac{z+3}{4}$ and parallel to the $xy$-plane is

The equation of the plane passing through the origin and containing the line $\frac{x - 1}{5} = \frac{y - 2}{4} = \frac{z - 3}{5}$ is: