Amit M Agarwal Solutions for Chapter: Three Dimensional Geometry, Exercise 4: Work Book Exercise 14.4

The point of intersection of the line $\frac{x-1}{3}=\frac{y+2}{4}=\frac{z-3}{-2}$ and plane $2x-y+3z-1=0$, is

If equation of 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

The distance of the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and the plane $x-y+z=5$ from the point $\left(-1,-5,-10\right)$, is

The equation of the line passing through the point $\left(3,0,-4\right)$ and perpendicular to the plane $2x-3y+5z-7=0$, is

The equation of the plane passing through the point $\left(2,-1,-3\right)$ and parallel to the lines $\frac{x-1}{3}=\frac{y+2}{2}=\frac{z}{-4}$ and $\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2},$ is

The equation of line of intersection of planes $4x+4y-5z=12$ and $8x+12y-13z=32$ can be written as

If the plane $3x+y+2z+5=0$ is parallel to the line $\frac{3x-1}{2b}=3-y=\frac{z-1}{a}$, then the value of $3a+3b$ is

The distance of the point $\left(3,8,2\right)$ from the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}$ measured parallel to the plane $3x+2y-2z+17=0$, is