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Lines (1-x)/3 = (7y-14)/2p = (z-3)/2 and (7-7x)/3p = (y-5)/1 = (6-z)/5 are perpendicular: find p

The same question, asked 2 times: 12 May 2024 S1*, 25 Apr 2025 S1. Every wording below, oldest first.

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  1. Question 1

    12 May 2024 S1

    147
    If the lines 1−x3=7y−142p=z−32\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2} and 7−7x3p=y−51=6−z5\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5} are at right angles, then P=P=

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  2. Question 2

    25 Apr 2025 S1

    125
    If the lines 1−x2=7y+42λ=2z−52\frac{1 -x}{2}=\frac{7y+ 4}{2\lambda}=\frac{2z- 5}{2} and 7−7x3λ=y−17=6−z5\frac{7 - 7x}{3\lambda}=\frac{y- 1}{7}=\frac{6 -z}{5} are at right angle, then the value of λ\lambda is

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