Mathematics

Mathematics

Problem

10. Find the values of pp so that the lines 1−x3=7y−142p=z−32\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2} and 7−7x3p=y−51=6−z5\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5} are at right angles.

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