Amit M Agarwal Solutions for Chapter: Three Dimensional Coordinate System, Exercise 11: Three Dimensional Coordinate System Exercise 8

Consider the lines $L_1:\frac{x-1}{2}=\frac{y}{-1}=\frac{z+3}{1}, L_2:\frac{x-4}{1}=\frac{y+3}{1}=\frac{z+3}{2}$ and the planes $P_1:7x+y+2z=3, P_2:3x+5y-6z=4$. Let $ax+by+cz=d$ be the equation of the plane passing through the point of intersection of lines $L_2$ and $L_1$ perpendicular to planes $P_1$ and $P_2$.

Match list $\mathrm{I}$ with list $\mathrm{II}$ and select the correct answer using the code given below the lists.

The distance of the point $\left(1,0,2\right)$ from the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and the plane $x–y+z=16$, is

The equation of the plane containing the line $2x–5y+z=3, x+y+4z=5$ and parallel to the plane $x+3y+6z=1$ is

The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and $l^2=m^2+n^2$ is

A line $AB$ in three- dimensional space makes angle $45°$ and $120°$ with the positive $X$-axis and the positive $Y$-axis, respectively. If $AB$ makes an acute angle $\theta$ with the positive $Z$-axis, then $\theta$ equals

If $\left(2,3,5\right)$ is one end of the diameter of the sphere $x^2+y^2+z^2–6x–12y–2z+20=0$, then the coordinates of the other end of the diameter are

The angle between the line $2x=3y=-z$ and $6x=-y=-4z$ is

The plane $x+2y-z=4$ cuts the sphere $x^2+y^2+z^2-x+z-2=0$ in a circle of radius