If the line $r = a + t b$ is parallel to the plane $r = c + l d + m e$,then

Let the plane $P$ contain the line $2x+y-z-3=0=5x-3y+4z+9$ and be parallel to the line $\frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5}$. Then the distance of the point $A(8,-1,-19)$ from the plane $P$ measured parallel to the line $\frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12}$ is equal to $............$.

The angle between the line of intersection of the two planes $r \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=5$ and $r \cdot(3 \hat{i}+3 \hat{j}-5 \hat{k})=3$,and the line $r=3 \hat{i}+2 \hat{j}+\hat{k}+t(5 \hat{i}+5 \hat{j}-7 \hat{k})$ is

The line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3y-\alpha z+\beta=0$,then the value of $\alpha^2+\alpha\beta+\beta^2$ is

Angle between the lines of intersection of the planes $x-y=0, 2x+y+z=0$ and $2x-z=0, x+y-3z=0$ is (in $^{\circ}$)

The equation of the plane,passing through the intersection of the planes $x+y+z=1$ and $2x+3y-z+4=0$ and parallel to $Y$-axis is