Given lines are

$$\frac{x-3}{4}=\frac{y-(-7)}{-11}=\frac{z-1}{5}$$ and

$$\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z-(-2)}{1}$$

Shortest distance between two lines,

$$\begin{aligned} & d=\frac{\left|\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)\right|}{\left|\left(\vec{b}_1 \times \vec{b}_2\right)\right|} \\ & \vec{a}_2-\vec{a}_1=2 \hat{i}+16 \hat{j}-3 \hat{k} \text { and } \\ & \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & -11 & 5 \\ 3 & -6 & 1 \end{array}\right|=19 \hat{i}+11 \hat{j}+9 \hat{k} \\ & \therefore \quad d=\frac{187}{\sqrt{563}} \end{aligned}$$