Find the values p so that line 31−x=2p7y−14=2z−3and3p7−7x=1y−5=56−z are at right angles.

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Find the values {p} so that line {{{1}-{x}}}{{3}}={{{7}{y}-{14

Question

Find the values $p$ so that line $\frac{1 - x}{3} = \frac{7 y - 14}{2 p} = \frac{z - 3}{2} a n d \frac{7 - 7 x}{3 p} = \frac{y - 5}{1} = \frac{6 - z}{5}$ are at right angles.

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Theodore Discussed

Find the values

p

so that line

\frac{1 - x}{3} = \frac{7 y - 14}{2 p} = \frac{z - 3}{2} a n d \frac{7 - 7 x}{3 p} = \frac{y - 5}{1} = \frac{6 - z}{5}

are at right angles.

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Question Text

Find the values

p

so that line

\frac{1 - x}{3} = \frac{7 y - 14}{2 p} = \frac{z - 3}{2} a n d \frac{7 - 7 x}{3 p} = \frac{y - 5}{1} = \frac{6 - z}{5}

are at right angles.