Question

The distance of the point (-1, -5, -10) 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 = 5, is
[AISSE 1985; DSSE 1984; MP PET 2002]

• A. 10
• B. 11
• C. 12
• D. 13

Answer 1

The correct option is D

13

Any point on the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=t$ is
(3t + 2, 4t - 1, 12t + 2)
This lies on x - y + z = 5
$\therefore$ 3t + 2 - 4t + 1 + 12t + 2 = 5 i.e., 11t = 0 $\Rightarrow$ t = 0
$\therefore$ Point is (2,-1,2). Its distance from (-1, -5, -10) is,
$=\sqrt{(2+1{)}^2+(-1+5{)}^2+(2+10{)}^2}=\sqrt{9+16+144}=13$.