jee-main 2025 Q89

jee-main · India · session2_04apr_shift2 Vectors: Lines & Planes Distance Computation (Point-to-Plane or Line-to-Line)
Q89. If the shortest distance between the lines $\frac { x - \lambda } { 3 } = \frac { y - 2 } { - 1 } = \frac { z - 1 } { 1 }$ and $\frac { x + 2 } { - 3 } = \frac { y + 5 } { 2 } = \frac { z - 4 } { 4 }$ is $\frac { 44 } { \sqrt { 30 } }$, then the largest possible value of $| \lambda |$ is equal to $\_\_\_\_$
Q89. If the shortest distance between the lines $\frac { x - \lambda } { 3 } = \frac { y - 2 } { - 1 } = \frac { z - 1 } { 1 }$ and $\frac { x + 2 } { - 3 } = \frac { y + 5 } { 2 } = \frac { z - 4 } { 4 }$ is $\frac { 44 } { \sqrt { 30 } }$, then the largest possible value of $| \lambda |$ is equal to $\_\_\_\_$\\