Verified Solution • 2025–26 Session moderate QJEEM22Mac87a: JEE Main 2022 Mathematics - Class 12 JEEM22Mac87a Solution: If the shortest distance between the lines and is , then the… If the shortest distance between the lines x−12=y−23=z−3λ{{x - 1} \over 2} = {{y - 2} \over 3} = {{z - 3} \over \lambda }2x−1=3y−2=λz−3 and x−21=y−44=z−55{{x - 2} \over 1} = {{y - 4} \over 4} = {{z - 5} \over 5}1x−2=4y−4=5z−5 is 13{1 \over {\sqrt 3 }}31 , then the sum of all possible value of λ\lambdaλ is : Multiple Choice Options A 6 B 12 C 15 D 16 Step-by-Step Verified Solution Let a→1=i^+2j^+3k^{\overrightarrow a _1} = \widehat i + 2\widehat j + 3\widehat ka1=i+2j+3k a→2=2i^+4j^+5k^{\overrightarrow a _2} = 2\widehat i + 4\widehat j + 5\widehat ka2=2i+4j+5k p→=2i^+3j^+λk^, q→=i^+4j^+5k^\overrightarrow p = 2\widehat i + 3\widehat j + \lambda \widehat k,\,\overrightarrow q = \widehat i + 4\widehat j + 5\widehat kp=2i+3j+λk,q=i+4j+5k ∴\therefore∴ p→×q→=(15−4λ)i^−(10−λ)j^+5k^\overrightarrow p \times \overrightarrow q = (15 - 4\lambda )\widehat i - (10 - \lambda )\widehat j + 5\widehat kp×q=(15−4λ)i−(10−λ)j+5k a→2−a→1=i^+2j^+2k^{\overrightarrow a _2} - {\overrightarrow a _1} = \widehat i + 2\widehat j + 2\widehat ka2−a1=i+2j+2k ∴\therefore∴ Shortest distance =∣(15−4λ)−2(10−λ)+10(15−4λ)2+(10−λ)2+25∣=13= \left| {{{(15 - 4\lambda ) - 2(10 - \lambda ) + 10} \over {\sqrt {{{(15 - 4\lambda )}^2} + {{(10 - \lambda )}^2} + 25} }}} \right| = {1 \over {\sqrt 3 }}=(15−4λ)2+(10−λ)2+25(15−4λ)−2(10−λ)+10=31 ⇒3(5−2λ)2=(15−4λ)2+(10−λ)2+25\Rightarrow 3{(5 - 2\lambda )^2} = {(15 - 4\lambda )^2} + {(10 - \lambda )^2} + 25⇒3(5−2λ)2=(15−4λ)2+(10−λ)2+25 ⇒5λ2−80λ+275=0\Rightarrow 5{\lambda ^2} - 80\lambda + 275 = 0⇒5λ2−80λ+275=0 ∴\therefore∴ Sum of values of λ=805=16\lambda = {{80} \over 5} = 16λ=580=16 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring