The equation of the plane through the intersection of the planes $x+2y+3z-4=0$ and $4x+3y+2z+1=0$ and passing through the origin is

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

Let the lines $l_1: \frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha}{-2}$ and $l_2: 3x+2y+z-2=0=x-3y+2z-13$ be coplanar. If the point $P(a, b, c)$ on $l_1$ is nearest to the point $Q(-4, -3, 2)$,then $|a|+|b|+|c|$ is equal to

The plane $2x - y + 3z + 5 = 0$ is rotated through $90^o$ about its line of intersection with the plane $5x - 4y - 2z + 1 = 0$. The equation of the plane in its new position is:

The distance of the point $(1, -5, 9)$ from the plane $x - y + z = 5$ measured along the line $x = y = z$ is

Let $\vec{r} \cdot (3\hat{i} - 2\hat{j} + 7\hat{k}) = 32$ be the equation of a plane. If the line with direction ratios $(5, b, 3)$ is parallel to the plane, then the value of $b$ is: