D.P. GUPTA, Preetima Bajpai and, Sanjeev Kumar Jha Solutions for Chapter: Three Dimensional Geometry, Exercise 1: Exercise 1

The equation of the plane through the line $x+y+z+3=0=2x-y+3z+1$ and parallel to the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ is

If the direction cosines of a line are $\left(\frac{1}{c},\frac{1}{c},\frac{1}{c}\right)$, then

The direction ratios of the intersection line of these planes $x-y+z-5=0,x-3y=0$ are

The line $\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-1}{-1}$ intersects the curve $xy=c^2, z=0$ if $c$ is equal to

The three planes $x+y=0,y+z=0$ and $x+z=0$

The equation of the sphere concentric with the sphere $x^2+y^2+z^2-2x-4y+6z+10=0$ and of double its surface area is $x^2+y^2+z^2-2x-4y+6z+k=0$, then $k=$

The shortest distance between the lines $\frac{x-7}{3}=\frac{y+4}{-16}=\frac{z-6}{7}$ and $\frac{x-10}{3}=\frac{y-30}{8}=\frac{4-z}{5}$ is

The shortest distance between the lines $x=y+2=6z-6$ and $x+1=2y=-12z$ is