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

$\begin{array}{1 1} p=\frac{70}{11} \\ p=\frac{-10}{11} \\ p=\frac{10}{7} \\p=\frac{11}{70} \end{array}$

Toolbox:
• Angle between the line is given by
• $\cos \theta=\bigg|\large\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt {{a_1}^2+{b_1}^2+{c_1}^2}\sqrt {{a_2}^2+{b_2}^2+{c_2}^2}}\bigg|$
Let $L_1$ be the line :$\large\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}$
$\large\frac{x-1}{-3}=\frac{y-2}{2p/7}=\frac{z-3}{2}$
Hence the Direction cosines of the vector $b_1$ in $L_1$ is $(-3,\large\frac{2p}{7}$$,2) Let this be (a_1,b_1,c_1) Let L_2 be the line :\large\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5} \large\frac{x-1}{-3p/7}=\frac{y-5}{1}=\frac{z-6}{-5} Hence the Direction cosines of the vector b_2 in L_2 is (-\large\frac{3p}{7}$$,1,-5)$
Let this be $(a_2,b_2,c_2)$
we know $\cos \theta=\bigg|\large\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt {{a_1}^2+{b_1}^2+{c_1}^2}\sqrt {{a_2}^2+{b_2}^2+{c_2}^2}}\bigg|$
Now substituting for $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$
$\cos \theta=\Bigg|\large\frac{-3 \times \Large\frac{-3p}{7}+\frac{2p}{7} \large \times 1+ 2 \times -5}{\sqrt {3^2+\bigg(\Large\frac{2p}{7}\bigg)^2+\large2^2}\sqrt {\bigg(\Large\frac{-3p}{7}\bigg)^2+\large1^2+(-5)^2}}\Bigg|$
But it is given $\theta=90^{\circ}$
Therefore $\cos 90=0$
$0=\Bigg|\large\frac{\Large\frac{9p}{7}+\frac{2p}{7}-\large10}{\sqrt {9+\Large\frac{4p^2}{49}+\large4}\sqrt{\Large\frac {9p^2}{7}+1+5}}\Bigg|$
=>$\bigg|\large\frac{9p}{7}+\frac{2p}{7}-$$10\bigg|=0 On simplifying we get, \bigg|\large\frac{11p}{7}\bigg|=$$10\qquad (ie) 11p=70$
$p=\large\frac{70}{11}$