Question

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

• A. $11$
• B. $12$
• C. $13$
• D. $14$

Answer 1

Answer: (C)

$13$

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

Plane$:x-y+z=5$

The points of line $L$ will be of the form $(x,y,z)=(3r+2,4r-1,12r+2)$ $(x,y,z)$ must satisfy plane equation.

$\therefore 3r+2-(4r-1)+12r+2=5r=0$

$(x,y,z)=(2,-1,2)$ is point of intersection of $L$ and plane.

$\therefore$ distance from $P(-1,-5,-10)$ can be found by distance formula.

$\therefore$ distance$=\sqrt{{(2+1)}^2+{(-1+5)}^2+{(2+10)}^2}=13$