The line $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is parallel to the plane

The distance of the point $(7,5,2)$ from the plane $3x+4y+z-8=0$ measured parallel to the line $\frac{x-1}{3}=\frac{y-2}{6}=\frac{z+1}{2}$ is:

The equation of the plane that contains the line of intersection of the planes $x+2y+3z-4=0$ and $2x+y-z+5=0$ and is perpendicular to the plane $5x+3y-6z+8=0$ is

The foot of the perpendicular drawn from the origin to the plane $2x - 3y + 4z = 29$ is:

The equation of the plane passing through the line $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{4}$ and perpendicular to the plane $x+2y+z=12$ is given by $ax+by+cz+4=0$. Then:

For real numbers $\alpha$ and $\beta \neq 0$,if the point of intersection of the straight lines $\frac{x-\alpha}{1}=\frac{y-1}{2}=\frac{z-1}{3}$ and $\frac{x-4}{\beta}=\frac{y-6}{3}=\frac{z-7}{3}$ lies on the plane $x+2y-z=8$,then $\alpha-\beta$ is equal to: