jee-main 2021 Q89

jee-main · India · session4_31aug_shift1 Vectors 3D & Lines Line-Plane Intersection
The square of the distance of the point of intersection of the line $\frac { x - 1 } { 2 } = \frac { y - 2 } { 3 } = \frac { z + 1 } { 6 }$ and the plane $2 x - y + z = 6$ from the point $( - 1 , - 1,2 )$ is
The square of the distance of the point of intersection of the line $\frac { x - 1 } { 2 } = \frac { y - 2 } { 3 } = \frac { z + 1 } { 6 }$ and the plane $2 x - y + z = 6$ from the point $( - 1 , - 1,2 )$ is