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

$\frac{70}{11}$

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

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

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

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

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

$\vec{v}_2=-\frac{3 m}{7} \hat{i}+\hat{j}-5 \hat{k}$

so  $\vec{v}_1 . \vec{v}_2 = 0$    (for right angles)

$\vec{v_1} . \vec{v_2}=\left(-3 \hat{i}+\frac{2 m}{7} \hat{j}+2 \hat{k}\right)\left(-\frac{3 m}{7} \hat{i}+\hat{j}-5 \hat{k}\right)=0$

so   $\frac{9 m}{7}+\frac{2 m}{7}-10=0$

$\Rightarrow \frac{11 m}{7}=10$

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