Thank you for reporting, we will resolve it shortly
Q.
If the lines $\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2}$ and $\frac{7-7 x}{3 p}=\frac{5-y}{1}=\frac{6-z}{5}$ are orthogonal to each other, then the value of $p$ is
Vector Algebra
Solution:
The first lines is $\frac{x-1}{-3}=\frac{y-2}{2 p / 7}=\frac{z-3}{2}$; parallel to ${V}_1=-3 \hat{i}+\frac{2 p}{7} \hat{j}+2 \hat{k}$
Second line is $\frac{(x-1)}{-3 p / 7}=\frac{y-5}{-1}=\frac{z-6}{-5}$; parallel to ${v}_2=-\frac{3 p}{7} \hat{i}-\hat{j}-5 \hat{k}$
Now, $-3 \cdot\left(\frac{-3 p }{7}\right)+\left(\frac{2 p }{7}\right) \cdot(-1)+2(-5)=0$
$\therefore 7 p =70 \Rightarrow p =10$