Find the angle between the lines $\vec{r}=3 \hat{i}-2 \hat{j}+6 \hat{k}+\lambda(2 \hat{i}+\hat{j}+2 \hat{k})$ and $\vec{r}=(2 \hat{j}-5 \hat{k})+\mu(6 \hat{i}+3 \hat{j}+2 \hat{k})$.

(N/A) The given lines are $\vec{r}=\vec{a}_{1}+\lambda \vec{b}_{1}$ and $\vec{r}=\vec{a}_{2}+\mu \vec{b}_{2}$.
Here,$\vec{b}_{1}=2 \hat{i}+\hat{j}+2 \hat{k}$ and $\vec{b}_{2}=6 \hat{i}+3 \hat{j}+2 \hat{k}$.
The angle $\theta$ between the lines is given by $\cos \theta = \frac{|\vec{b}_{1} \cdot \vec{b}_{2}|}{|\vec{b}_{1}| |\vec{b}_{2}|}$.
Calculating the dot product: $\vec{b}_{1} \cdot \vec{b}_{2} = (2)(6) + (1)(3) + (2)(2) = 12 + 3 + 4 = 19$.
Calculating the magnitudes: $|\vec{b}_{1}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4+1+4} = \sqrt{9} = 3$.
$|\vec{b}_{2}| = \sqrt{6^2 + 3^2 + 2^2} = \sqrt{36+9+4} = \sqrt{49} = 7$.
Thus,$\cos \theta = \frac{19}{3 \times 7} = \frac{19}{21}$.
Therefore,$\theta = \cos^{-1} \left( \frac{19}{21} \right)$.