Solution:

Let’s consider $\theta$ be the angle between the given lines.
If $\theta$ is the acute angle between $\vec{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}$ and $\vec{r}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}}$ then
$\cos \theta=\left|\frac{\overrightarrow{b_{1}} \overrightarrow{b_{2}}}{\left|\overrightarrow{b_{1}}\right|\left|\overrightarrow{b_{2}}\right|}\right|_{\ldots \ldots \text { (1) }}$

(i) $\vec{r}=2 \hat{i}-5 \hat{j}+\hat{k}+\lambda(3 \hat{i}+2 \hat{j}+6 \hat{k})$ and
$\vec{r}=7 \hat{i}-6 \hat{k}+\mu(\hat{i}+2 \hat{j}+2 \hat{k})$
Here $\overrightarrow{b_{1}}=3 \hat{i}+2 \hat{j}+6 \hat{k}$ and $\overrightarrow{b_{2}}=\hat{i}+2 \hat{j}+2 \hat{k}$
From eq. (1), we get
$\cos \theta=\left|\frac{(3 \hat{i}+2 \hat{j}+6 \hat{k}) \cdot(\hat{i}+2 \hat{j}+2 \hat{k})}{|3 \hat{i}+2 \hat{j}+6 \hat{k}| \cdot|\hat{i}+2 \hat{j}+2 \hat{k}|}\right|_{\ldots}$
It is known to us that,
$|\mathrm{ai}+\mathrm{b} \hat{\mathrm{j}}+\mathrm{ck}|=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}}$
Therefore,
$|3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}|=\sqrt{3^{2}+2^{2}+6^{2}}=\sqrt{9+4+36}=\sqrt{49}=7$
And
$|\hat{i}+2 \hat{j}+2 \hat{k}|=\sqrt{1^{2}+2^{2}+2^{2}}=\sqrt{1+4+4}=\sqrt{9}=3$
Now, it is known that
$\left(a_{1} \hat{i}+b_{1} \hat{j}+c_{1} \hat{k}\right) \cdot\left(a_{2} \hat{i}+b_{2} \hat{j}+c_{2} \hat{k}\right)=a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}$
Therefore,
$(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})=3 \times 1+2 \times 2+6 \times 2=3+4+12=19$
By eq.(2), we get
$\cos \theta=\frac{19}{7 \times 3}=\frac{19}{21}$
$\theta=\cos ^{-1}\left(\frac{19}{21}\right)$
So,
$(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})=3 \times 1+2 \times 2+6 \times 2=3+4+12=19$
From eq. (2), we get
$\begin{array}{l}
\cos \theta=\frac{19}{7 \times 3}=\frac{19}{21} \\
\theta=\cos ^{-1}\left(\frac{19}{21}\right)
\end{array}$

(ii) $\vec{r}=3 \hat{i}+\hat{j}-2 \hat{k}+\lambda(\hat{i}-\hat{j}-2 \hat{k})$ and
$\vec{r}=2 \hat{i}-\vec{j}-56 \hat{k}+\mu(3 \hat{i}-5 \hat{j}-4 \hat{k})$
Here, $\overrightarrow{b_{1}}=\hat{i}-\hat{j}-2 \hat{k}$ and $\overrightarrow{b_{2}}=3 \hat{i}-5 \hat{j}-4 \hat{k}$
Therefore, from eq.(1), we have
$\cos \theta=\left|\frac{(\hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \cdot(3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-4 \hat{\mathrm{k}})}{|\hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}} \| 3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}|}\right|_{..$
It is known to us that,
$|\mathrm{a} \hat{\mathrm{i}}+\mathrm{b} \hat{\mathrm{j}}+\mathrm{ck}|=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}}$
Therefore,
$|\hat{i}-\hat{j}-2 \hat{k}|=\sqrt{1^{2}+(-1)^{2}+2^{2}}=\sqrt{1+1+4}=\sqrt{6}=\sqrt{3} \times \sqrt{2}$
And
$|3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}|=\sqrt{3^{2}+(-5)^{2}+(-4)^{2}}=\sqrt{9+25+16}=\sqrt{50}=5 \sqrt{2}$
Now, it is known that
$\left(a_{1} \hat{i}+b_{1} \hat{j}+c_{1} \hat{k}\right) \cdot\left(a_{2} \hat{i}+b_{2} \hat{j}+c_{2} \hat{k}\right)=a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}$
$\therefore(\hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \cdot(3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-4 \hat{\mathrm{k}})=1 \times 3+(-1) \times(-5)+(-2) \times(-4)=3+5+8=16$
By eq.(3), we get
$\begin{array}{l}
\cos \theta=\frac{16}{\sqrt{3} \times \sqrt{2} \times 5 \sqrt{2}}=\frac{16}{5 \times 2 \sqrt{3}}=\frac{8}{5 \sqrt{3}} \\
\theta=\cos ^{-1}\left(\frac{8}{5 \sqrt{3}}\right) \\
\theta=\cos ^{-1}\left(\frac{8}{5 \sqrt{3}}\right)
\end{array}$