The value of $\lambda$, so that the lines $\frac{1-x}{3}=\frac{7 y-14}{2 \lambda}=\frac{z-3}{2}$ and $\frac{7-7 x}{3 \lambda}=\frac{y-5}{1}=\frac{6-z}{5}$ are perpendicular is:

$\frac{70}{11}$

$l_1: \frac{1-x}{3}=\frac{7 y-14}{2 \lambda}=\frac{z-3}{2}$

$l_2: \frac{7-7 x}{3 \lambda}=\frac{4-5}{1}=\frac{6-z}{5}$

$l_1:=\frac{x-1}{-3}=\frac{y-2}{\frac{2 \lambda}{7}}=\frac{z-3}{z}$

$l_2: \frac{x-1}{\frac{-3 \lambda}{7}}=\frac{y-5}{1}=\frac{z-6}{-5}$

so vector

$\vec{v}_1=-3 \hat{i}+\frac{2 \lambda}{7} \hat{j}+2 \hat{k}$

$\vec{v}_1 || l_1$

veetor

$\vec{v_2}=-\frac{3 \lambda}{7} \hat{i}+\hat{j}-5 \hat{k}$

$\vec{v}_2 || l_2$

 for $l_1 \perp l_2$

$\Rightarrow \vec{v}_1 \perp \vec{v}_2 \Rightarrow \vec{v}_1 . \vec{v}_2=0$

$\Rightarrow\left(-3 \hat{i}+\frac{2 \lambda}{7} \hat{j}+2 \hat{k}\right) . \left(-\frac{3 \lambda}{7} \hat{i}+\hat{j}-5 \hat{k}\right)=0$

$\frac{9 \lambda}{7}+\frac{2 \lambda}{7}-10=0 \Rightarrow \frac{11 \lambda}{7}=10$

$\lambda=\frac{70}{11}$