# 12.   Find the values of p so that the lines $\frac{1-x}{3}=\frac{7y-14}{2p}= \frac{z-3}{2}$ and $\frac{7-7x}{3p}=\frac{y-5}{1}= \frac{6-z}{5}$ are at right angles.

First we have to write the given equation of lines in the standard form;

$\frac{x-1}{-3}=\frac{y-2}{\frac{2p}{7}}= \frac{z-3}{2}$   and  $\frac{x-1}{\frac{-3p}{7}}=\frac{y-5}{1}= \frac{z-6}{-5}$

Then we have the direction ratios of the above lines as;

$-3,\ \frac{2p}{7},\ 2$   and   $\frac{-3p}{7},\ 1,\ -5$   respectively..

Two lines with direction ratios $a_{1},b_{1},c_{1}$ and $a_{2},b_{2},c_{2}$  are perpendicular to each other if,  $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}= 0$

$\therefore (-3).\left ( \frac{-3p}{7} \right )+(\frac{2p}{7}).(1) + 2.(-5) = 0$

$\Rightarrow \frac{9p}{7}+ \frac{2p}{7} =10$

$\Rightarrow 11p =70$

$\Rightarrow p =\frac{70}{11}$

Thus, the value of p is $\frac{70}{11}$.

## Related Chapters

### Preparation Products

##### Knockout KCET 2021

An exhaustive E-learning program for the complete preparation of KCET exam..

₹ 4999/- ₹ 2999/-
##### Knockout KCET JEE Main 2021

It is an exhaustive preparation module made exclusively for cracking JEE & KCET.

₹ 27999/- ₹ 16999/-
##### Knockout NEET Sept 2020

An exhaustive E-learning program for the complete preparation of NEET..

₹ 15999/- ₹ 6999/-
##### Rank Booster NEET 2020

This course will help student to be better prepared and study in the right direction for NEET..

₹ 9999/- ₹ 4999/-