If the shortest distance between the straight lines \(3(x-1)=6(y-2)=2(z-1)\) and \(4(x-2)=2(y-\lambda)=(z-3), \lambda \in R\) is \(\frac{1}{\sqrt{38}}\), then the integral value of \(\lambda\) is equal to:

[JEE Main 2021, 22 Jul (Shift 2)]

Question

If the shortest distance between the straight lines \(3(x-1)=6(y-2)=2(z-1)\) and \(4(x-2)=2(y-\lambda)=(z-3), \lambda \in R\) is \(\frac{1}{\sqrt{38}}\), then the integral value of \(\lambda\) is equal to:

[JEE Main 2021, 22 Jul (Shift 2)]

YESPYQ · Accepted Answer

Options are free to see. Unlock the correct answer and full explanation with PYQ Pass.

If the shortest distance between the straight lines \(3(x-1)=6(y-2)=2(z-1)\) and \(4(x-2)=2(y-\lambda)=(z-3), \lambda \i…

Practice more JEE Maths PYQs