jee-main 2025 Q90

jee-main · India · session2_02apr_shift1 Vectors: Lines & Planes Distance Computation (Point-to-Plane or Line-to-Line)
Q90. If the shortest distance between the lines $\frac { x + 2 } { 2 } = \frac { y + 3 } { 3 } = \frac { z - 5 } { 4 }$ and $\frac { x - 3 } { 1 } = \frac { y - 2 } { - 3 } = \frac { z + 4 } { 2 }$ is $\frac { 38 } { 3 \sqrt { 5 } } \mathrm { k }$, and $\int _ { 0 } ^ { \mathrm { k } } \left[ x ^ { 2 } \right] \mathrm { d } x = \alpha - \sqrt { \alpha }$, where $[ x ]$ denotes the greatest integer function, then $6 \alpha ^ { 3 }$ is equal to $\_\_\_\_$墐
ANSWER KEYS

\begin{tabular}{|l|l|} \hline 1. (3) & 2. (1) \hline 9. (3) & 10. (3) \hline 17. (1) & 18. (2) \hline 25. (16) & 26. (1) \hline 33. (4) & 34. (3) \hline 41. (4) & 42. (1) \hline 49. (4) & 50. (1) \hline 57. (100) & 58. (4) \hline 65. (1) & 66. (3) \hline
Q90. If the shortest distance between the lines $\frac { x + 2 } { 2 } = \frac { y + 3 } { 3 } = \frac { z - 5 } { 4 }$ and $\frac { x - 3 } { 1 } = \frac { y - 2 } { - 3 } = \frac { z + 4 } { 2 }$ is $\frac { 38 } { 3 \sqrt { 5 } } \mathrm { k }$, and $\int _ { 0 } ^ { \mathrm { k } } \left[ x ^ { 2 } \right] \mathrm { d } x = \alpha - \sqrt { \alpha }$, where $[ x ]$ denotes the greatest integer function, then $6 \alpha ^ { 3 }$ is equal to $\_\_\_\_$墐

\section*{ANSWER KEYS}
\begin{center}
\begin{tabular}{|l|l|}
\hline
1. (3) & 2. (1) \\
\hline
9. (3) & 10. (3) \\
\hline
17. (1) & 18. (2) \\
\hline
25. (16) & 26. (1) \\
\hline
33. (4) & 34. (3) \\
\hline
41. (4) & 42. (1) \\
\hline
49. (4) & 50. (1) \\
\hline
57. (100) & 58. (4) \\
\hline
65. (1) & 66. (3) \\
\hline