If the point of intersection of the lines $\frac{x+1}{3} = \frac{y+a}{5} = \frac{z+b+1}{7}$ and $\frac{x-2}{1} = \frac{y-b}{4} = \frac{z-2a}{7}$ lies on the $xy$-plane, then the value of $a+b$ is:

If the line $\frac{x - 3}{2} = \frac{y + 2}{-1} = \frac{z + 4}{3}$ lies in the plane $lx + my - z = 9$,then $l^2 + m^2 = \dots$

If the line $\bar{r}=(\hat{i}-2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+\hat{j}+2 \hat{k})$ is parallel to the plane $\bar{r} \cdot(3 \hat{i}-2 \hat{j}-m \hat{k})=5$,then the value of $m$ is:

Let the plane containing the line of intersection of the planes $P_1: x+(\lambda+4)y+z=1$ and $P_2: 2x+y+z=2$ pass through the points $(0,1,0)$ and $(1,0,1)$. Then the distance of the point $(2\lambda, \lambda, -\lambda)$ from the plane $P_2$ is (in $\sqrt{6}$)

Let $P$ be a plane $lx + my + nz = 0$ containing the line $\frac{1-x}{1} = \frac{y+4}{2} = \frac{z+2}{3}$. If plane $P$ divides the line segment $AB$ joining points $A(-3, -6, 1)$ and $B(2, 4, -3)$ in the ratio $k : 1$,then the value of $k$ is equal to

The angle between the line $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z + 3}{-2}$ and the plane $x + y + 4 = 0$ is ......... $^o$.