Question

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

• A. 13
• B. 11
• C. 12
• D. none of these

Answer 1

The correct option is A 13

Line equation

$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=\lambda$

Any point $(x,y,z)$ on line will be $(3\lambda +2,4\lambda -1,12\lambda +2)$

point of intersection of line with plane

$x-y+z=5$

$3\lambda +2-4\lambda +1+12\lambda +2=15$

$\Rightarrow 11\lambda +5=5$

$\Rightarrow 11\lambda =0$

$\Rightarrow \lambda =0$

$\therefore$ Point of $x_n$ is $(2,-1,2)$

Distance between $(-1,-5,-10)$ and $(2,-1,2)$ is

$\sqrt{(2+1{)}^2+(-1+5{)}^2+(2+10{)}^2}=\sqrt{3^2+4^2+{12}^2}$

$=\sqrt{5^2+{12}^2}$

$=\sqrt{{13}^2}$

$=13$.