Q. 47 If the line \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-\mathrm{z}}{5} are at right angles, then \mathrm{p}=
A. \frac{70}{11}
B. \frac{11}{70}
C. \frac{-70}{11}
D. \frac{-11}{70}
Answer: A
Solution:
Given lines can be written as
\frac{x-1}{-3}=\frac{y-2}{\frac{2 p}{7}}=\frac{z-3}{2} \text { and } \frac{x-1}{-\frac{3 p}{7}}=\frac{y-5}{1}=\frac{z-6}{-5}As these lines are at right angles, we get
(-3)\left(-\frac{3 p}{7}\right)+\left(\frac{2 p}{7}\right)(1)+(2)(-5)=0 \therefore \quad \frac{9 p}{7}+\frac{2 p}{7}-10=0 \therefore \quad \frac{11 p}{7}=10 \Rightarrow p=\frac{70}{11}