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 ________.

JEE Main 2024 · MathsNumerical answerNumerical valueMediumCalculation

JEE Main 6 April 2024, Shift 2, Maths Q28

Question 28 of 90 in this shift, Maths question 28 of 30, Section B.

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}

\frac{44}{\sqrt{30}}

, then the largest possible value of

|\lambda|

is equal to ________.

Official answer

NTA final key.