The point of intersection of the lines $\frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}\text{and}\frac{x+3}{-36}=\frac{y-3}{2}=\frac{z-6}{4}$ is (A) $(21,\frac{5}{3},\frac{10}{3})$ (B) $(2,10,4)$ (C) $(-3,3,6)$ (D) $(5,7,-2)$

The point of intersection of lines $\frac{x-5}{9}=\frac{y-7}{-1}=\frac{z+2}{1}$ and $\frac{x+3}{-36}=\frac{y-3}{2}=\frac{z-6}{4}$ is

The point of intersection of the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-4}{5}=\frac{y-1}{2}=z$ is

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" The point of intersection of lines "(x-1)/(2)=(y-2)/(3)=(z-3)/(4)" and "(x-4)/(5)=(y-1)/(2)=(z)/(1)" is "

The coordinates of the point of intersection of the lines $\frac{x-1}{1}=\frac{y+2}{3}=\frac{z-2}{-2}$ with the plane $3x+4y+5z-25=0$ is (A) $(5,6,-10)$ (B) $(5,10,-6)$ (C) $(-6,5,10)$ (D) $(-6,10,5)$

The lines $\frac{x-1}{2}=\frac{y-2}{4}=\frac{z-3}{6}$ and $\frac{x-1}{4}=\frac{y-2}{5}=\frac{z-3}{7}$ are

The distance of point of intersection of lines $\frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7}\text{and}\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$ from (1, -4,7), is

Prove that the lines $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}and\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-6}{7}$ are coplanar . Also, find the plane containing these two lines.

The distance of point of intersection of lines $\frac{x-4}{1}=\frac{x+3}{-4}=\frac{z-1}{7}\text{and}\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$ from $(1,-4,7)$ is

Find the angle between the lines
$\frac{5-x}{3}=\frac{y+3}{-4},z=7\text{and}\frac{x}{1}=\frac{1-y}{2}=\frac{z-6}{2}$