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The value of λ\lambda, so that the lines 1−x3=7y−142λ=z−32\frac{1-x}{3} = \frac{7y-14}{2\lambda} = \frac{z-3}{2} and 7−7x3λ=y−51=6−z5\frac{7-7x}{3\lambda} = \frac{y-5}{1} = \frac{6-z}{5} are perpendicular is:

Solution

✅ Correct Option: 2

Direction ratios of line 1: (−3,2λ/7,2)(-3, 2\lambda/7, 2). Direction ratios of line 2: (−3λ/7,1,−5)(-3\lambda/7, 1, -5). Perpendicularity requires dot product zero: (−3)(−3λ/7)+(2λ/7)(1)+(2)(−5)=0(-3)(-3\lambda/7) + (2\lambda/7)(1) + (2)(-5) = 0, i.e. 9λ/7+2λ/7−10=09\lambda/7 + 2\lambda/7 - 10 = 0, so 11λ/7=1011\lambda/7 = 10, giving λ=70/11\lambda = 70/11.

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