The equation of the plane passing through the origin and perpendicular to the line $x = 2y = 3z$ is

Equation of the plane which passes through the point of intersection of lines $\frac{x - 1}{3} = \frac{y - 2}{1} = \frac{z - 3}{2}$ and $\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ and has the largest distance from the origin is

If the line $\frac{x - 3}{1} = \frac{y + 2}{-1} = \frac{z + \lambda}{-2}$ lies in the plane $2x - 4y + 3z = 2$,then the shortest distance between this line and the line $\frac{x - 1}{12} = \frac{y}{9} = \frac{z}{4}$ 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 image of the point $(3, 2, 1)$ in the plane $2x - y + 3z = 7$ is

If the foot of the perpendicular from the point $A(-1, 4, 3)$ on the plane $P: 2x + my + nz = 4$ is $B\left(-2, \frac{7}{2}, \frac{3}{2}\right)$,then the distance of the point $A$ from the plane $P$,measured parallel to a line with direction ratios $3, -1, -4$,is equal to.