If the line $\frac{1-x}{3}=\frac{7 y-14}{2 \lambda}=\frac{z-3}{2}$ and $\frac{7-7 x}{3…
If the line $\frac{1-x}{3}=\frac{7 y-14}{2 \lambda}=\frac{z-3}{2}$ and $\frac{7-7 x}{3 \lambda}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles, then $\lambda=$
$\frac{-70}{11}$
$\frac{70}{11}$
$\frac{11}{70}$
$\frac{-11}{70}$
Solution
We have lines
$\frac{x-1}{-3}=\frac{7(y-2)}{2 \lambda}=\frac{z-3}{2} \text { and } \frac{7(1-x)}{3 \lambda}=\frac{y-5}{1}=\frac{z-6}{-5}$
i.e. $\frac{x-1}{-3}=\frac{y-2}{\left(\frac{2 \lambda}{7}\right)}=\frac{z-3}{2}$ and $\frac{x-1}{\left(\frac{-3 \lambda}{7}\right)}=\frac{y-5}{1}=\frac{z-6}{5}$
Since given lines are at right angles, we write
$\begin{aligned}
& (-3)\left(\frac{-3 \lambda}{7}\right)+\left(\frac{2 \lambda}{7}\right)(1)+(2)(-5)=0 \\
& \therefore \frac{9 \lambda}{7}+\frac{2 \lambda}{7}-10=0 \Rightarrow 11 \lambda=70 \Rightarrow \lambda=\frac{70}{11}
\end{aligned}$