jee-main 2025 Q90

jee-main · India · session2_03apr_shift2 Vectors: Lines & Planes Distance Computation (Point-to-Plane or Line-to-Line)
Q90. Let the point $( - 1 , \alpha , \beta )$ lie on the line of the shortest distance between the lines $\frac { x + 2 } { - 3 } = \frac { y - 2 } { 4 } = \frac { z - 5 } { 2 }$ and $\frac { x + 2 } { - 1 } = \frac { y + 6 } { 2 } = \frac { z - 1 } { 0 }$. Then $( \alpha - \beta ) ^ { 2 }$ is equal to $\_\_\_\_$
ANSWER KEYS

\begin{tabular}{|l|l|l|l|} \hline 1. (3) & 2. (3) & 3. (4) & 4. (2) \hline 9. (3) & 10. (4) & 11. (3) & 12. (2) \hline 17. (2) & 18. (2) & 19. (3) & 20. (2) \hline 25. (16) & 26. (5) & 27. (500) & 28. (4) \hline 
Q90. Let the point $( - 1 , \alpha , \beta )$ lie on the line of the shortest distance between the lines $\frac { x + 2 } { - 3 } = \frac { y - 2 } { 4 } = \frac { z - 5 } { 2 }$ and $\frac { x + 2 } { - 1 } = \frac { y + 6 } { 2 } = \frac { z - 1 } { 0 }$. Then $( \alpha - \beta ) ^ { 2 }$ is equal to $\_\_\_\_$

\section*{ANSWER KEYS}
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
1. (3) & 2. (3) & 3. (4) & 4. (2) \\
\hline
9. (3) & 10. (4) & 11. (3) & 12. (2) \\
\hline
17. (2) & 18. (2) & 19. (3) & 20. (2) \\
\hline
25. (16) & 26. (5) & 27. (500) & 28. (4) \\
\hline
Paper Questions
Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 Q21 Q22 Q24 Q25 Q26 Q27 Q28 Q29 Q30 Q31 Q32 Q33 Q34 Q35 Q36 Q37 Q38 Q39 Q40 Q41 Q42 Q43 Q44 Q45 Q46 Q47 Q49 Q50 Q51 Q52 Q53 Q54 Q55 Q56 Q57 Q58 Q59 Q60 Q61 Q62 Q63 Q64 Q65 Q66 Q67 Q68 Q69 Q71 Q72 Q73 Q74 Q75 Q76 Q77 Q78 Q79 Q80 Q81 Q82 Q83 Q84 Q85 Q87 Q88 Q89 Q90