jee-main 2023 Q89

jee-main · India · session2_08apr_shift1 Vectors 3D & Lines Distance from a Point to a Line (Show/Compute)
Let $\lambda _ { 1 } , \lambda _ { 2 }$ be the values of $\lambda$ for which the points $\left( \frac { 5 } { 2 } , 1 , \lambda \right)$ and $( - 2 , 0 , 1 )$ are at equal distance from the plane $2 x + 3 y - 6 z + 7 = 0$. If $\lambda _ { 1 } > \lambda _ { 2 }$, then the distance of the point $\left( \lambda _ { 1 } - \lambda _ { 2 } , \lambda _ { 2 } , \lambda _ { 1 } \right)$ from the line $\frac { x - 5 } { 1 } = \frac { y - 1 } { 2 } = \frac { z + 7 } { 2 }$ is $\_\_\_\_$
Let $\lambda _ { 1 } , \lambda _ { 2 }$ be the values of $\lambda$ for which the points $\left( \frac { 5 } { 2 } , 1 , \lambda \right)$ and $( - 2 , 0 , 1 )$ are at equal distance from the plane $2 x + 3 y - 6 z + 7 = 0$. If $\lambda _ { 1 } > \lambda _ { 2 }$, then the distance of the point $\left( \lambda _ { 1 } - \lambda _ { 2 } , \lambda _ { 2 } , \lambda _ { 1 } \right)$ from the line $\frac { x - 5 } { 1 } = \frac { y - 1 } { 2 } = \frac { z + 7 } { 2 }$ is $\_\_\_\_$