1.If the lines 1−x2=y−8λ=z−52\frac{1-x}{2} = \frac{y-8}{\lambda} = \frac{z-5}{2}21−x=λy−8=2z−5 and x−115=y−33=z−11\frac{x-11}{5} = \frac{y-3}{3} = \frac{z-1}{1}5x−11=3y−3=1z−1 are perpendicular, then λ=\lambda =λ=a.4b.-4c.83\frac{8}{3}38d.−83-\frac{8}{3}−38Login to continueOnly logged in users canattempt or see the solution.
