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

• A. 10
• B. 11
• C. 12
• D. 13

Answer 1

Answer: (D)

13

Answer 2

Any point on the line $3 ( - 2 m - 3 n ) m + m n - 4 ( - 2 m - 3 n ) n = 0$ is

$- 6 m ^ { 2 } - 9 m n + m n + 8 m n + 12 n ^ { 2 } = 0$

This lies on $6 m ^ { 2 } - 12 n ^ { 2 } = 0$

$m ^ { 2 } - 2 n ^ { 2 } = 0$ $m + \sqrt { 2 } n = 0$ i.e., $m - \sqrt { 2 } n = 0$

$\therefore$Point is $( 2 , - 1,2 )$ . Its distance from $( - 1 , - 5 , - 10 )$ is,

=$\sqrt { ( 2 + 1 ) ^ { 2 } + ( - 1 + 5 ) ^ { 2 } + ( 2 + 10 ) ^ { 2 } }$= $\sqrt { 9 + 16 + 144 } = 13$ .