Question

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

• A. $13$
• B. $12$
• C. $11$
• D. $8$

Answer 1

The correct option is A $13$

$3r+2,4r-1,12r+2$

$x-y+z=5$

$3r+2-4r+1+12r+2=5$

$12r=0r=0$

$(2,-1,2)Q(-1,-5,-10)$

$=\left(\sqrt{3^2+4^2+(12{)}^2}\right)$

$=\sqrt{169}$

$=13$