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The values of p for which the lines $\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2}$ and $\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles is: |
$\frac{70}{11}$ $\frac{11}{70}$ $\frac{7}{11}$ $\frac{11}{7}$ |
$\frac{70}{11}$ |
The correct answer is Option (1) → $\frac{70}{11}$ |
