The plane $3x + 4y + 6z + 7 = 0$ is rotated about the line $r = (\hat{i} + 2\hat{j} - 3\hat{k}) + t(2\hat{i} - 3\hat{j} + \hat{k})$ until the plane passes through the origin. The equation of the plane in the new position is

The equation of the plane passing through the point of intersection of the planes $2x-y+z-3=0$ and $4x-3y+5z+9=0$ and parallel to the line $\frac{x+1}{2}=\frac{y+3}{4}=\frac{z-3}{5}$ is $\alpha x+\beta y+\gamma z+d=0$. Then $\alpha+\beta+\gamma+d=$

If the equation of the plane passing through the mirror image of a point $(2,3,1)$ with respect to the line $\frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1}$ and containing the line $\frac{x-2}{3}=\frac{1-y}{2}=\frac{z+1}{1}$ is $\alpha x+\beta y+\gamma z=24$,then $\alpha+\beta+\gamma$ is equal to ..... .

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

Find the position vector of the point where the line $r = (i - j + k) + t(i + j + k)$ meets the plane $r \cdot (i + j - k) = 5$.

$A$ plane makes positive intercepts of unit length on each of $X$ and $Y$ axes. If it passes through the point $(-1, 1, 2)$ and makes an angle $\theta$ with the $X$-axis,then $\theta$ is