Since, $\overrightarrow{O A}$ is inclined at $60^{\circ}$ to $O X$ and at $45^{\circ}$ to $O Y$. Let $\overrightarrow{O A}$ makes angle $\alpha$ with $O Z$.

$$\therefore \quad \cos ^2 60^{\circ}+\cos ^2 45^{\circ}+\cos ^2 \alpha=1$$

$\Rightarrow \quad\left(\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2+\cos ^2 \alpha=1 \quad\left[\because l^2+m^2+n^2=1\right]$

$$\begin{array}{lr} \Rightarrow & \frac{1}{4}+\frac{1}{2}+\cos ^2 \alpha=1 \\ \Rightarrow & \cos ^2 \alpha=1-\left(\frac{1}{2}+\frac{1}{4}\right) \\ \Rightarrow & \cos ^2 \alpha=1-\left(\frac{6}{8}\right) \\ \Rightarrow & \cos ^2 \alpha=\frac{1}{4} \\ \Rightarrow & \cos \alpha=\frac{1}{2}=\cos 60^{\circ} \end{array}$$

$$\begin{aligned} \therefore & =60^{\circ} \\ \therefore \quad \overrightarrow{O A} & =|\overrightarrow{O A}|\left(\frac{1}{2} \hat{\mathbf{i}}+\frac{1}{\sqrt{2}} \hat{\mathbf{j}}+\frac{1}{2} \hat{\mathbf{k}}\right) \\ & =10\left(\frac{1}{2} \hat{\mathbf{i}}+\frac{1}{\sqrt{2}} \hat{\mathbf{j}}+\frac{1}{2} \hat{\mathbf{k}}\right) \quad [\because|\overrightarrow{O A}|=10]\\ & =5 \hat{\mathbf{i}}+5 \sqrt{2} \hat{\mathbf{j}}+5 \hat{\mathbf{k}} \end{aligned}$$

Let $\overrightarrow{\mathbf{a}}=3 \hat{i}-2 \hat{j}+6 \hat{k}$ and $\overrightarrow{\mathbf{b}}=\hat{i}-2 \hat{j}+3 \hat{k}$

So, vector equation of the line, which is parallel to the vector $\vec{a}=3 \hat{i}-2 \hat{j}+6 \hat{k}$ and passes through the vector $\overrightarrow{\mathbf{b}}=\hat{i}-2 \hat{j}+3 \hat{k}$ is $\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{b}}+\lambda \overrightarrow{\mathbf{a}}$.

$$\begin{array}{l} \therefore & \overrightarrow{\mathrm{r}} =(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) \\ \Rightarrow & (x \hat{\mathrm{i}}+y \hat{\mathrm{j}}+z \hat{\mathrm{k}})-(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) =\lambda(33 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+6 \hat{\mathbf{k}}) \\ \Rightarrow &(x-1) \hat{\mathbf{i}}+(y+2) \hat{\mathbf{j}}+(z-3) \hat{\mathbf{k}} =\lambda(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) \end{array}$$

We have,

$$\begin{array}{ll} \text { We have, } & x_1=1, y_1=2, z_1=3 \text { and } a_1=2, b_1=3, c_1=4 \\ \text { Also, } & x_2=4, y_2=1, z_2=0 \text { and } a_2=5, b_2=2, c_2=1 \end{array}$$

If two lines intersect, then shortest distance between them should be zero.

$\therefore \quad$ Shortest distance between two given lines

$=\frac{\left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\end{array}\right|}{\sqrt{\left(b_1 c_2-b_2 c_1\right)^2+\left(c_1 a_2-c_2 a_1\right)^2+\left(a_1 b_2-a_2 b_1\right)^2}}$

$=\frac{\left|\begin{array}{ccc}4-1 & 1-2 & 0-3 \\ 2 & 3 & 4 \\ 5 & 2 & 1\end{array}\right|}{\sqrt{(3 \cdot 1-2 \cdot 4)^2+(4 \cdot 5-1 \cdot 2)^2+(2 \cdot 2-5 \cdot 3)^2}}$

$$\begin{aligned} & =\frac{\left|\begin{array}{ccc} 3 & -1 & -3 \\ 2 & 3 & 4 \\ 5 & 2 & 1 \end{array}\right|}{\sqrt{25+324+121}} \\ & =\frac{3(3-8)+1(2-20)-3(4-15)}{\sqrt{470}} \\ & =\frac{-15-18+33}{\sqrt{470}}=\frac{0}{\sqrt{470}}=0 \end{aligned}$$

Therefore, the given two lines are intersecting.

For finding their point of intersection for first line,

$$\begin{aligned} \frac{x-1}{2} & =\frac{y-2}{3}=\frac{z-3}{4}=\lambda \\ \Rightarrow \quad x & =2 \lambda+1, y=3 \lambda+2 \text { and } z=4 \lambda+3 \end{aligned}$$

Since, the lines are intersecting. So, let us put these values in the equation of another line.

$$\begin{array}{ll} \text { Thus, } & \frac{2 \lambda+1-4}{5}=\frac{3 \lambda+2-1}{2}=\frac{4 \lambda+3}{1} \\ \Rightarrow & \frac{2 \lambda-3}{5}=\frac{3 \lambda+1}{2}=\frac{4 \lambda+3}{1} \\ \Rightarrow & \frac{2 \lambda-3}{5}=\frac{4 \lambda+3}{1} \\ \Rightarrow & 2 \lambda-3=20 \lambda+15 \\ \Rightarrow & 18 \lambda=-18=-1 \end{array}$$

So, the required point of intersection is

$$\begin{aligned} & x=2(-1)+1=-1 \\ & y=3(-1)+2=-1 \\ & z=4(-1)+3=-1 \end{aligned}$$

Thus, the lines intersect at $(-1,-1,-1)$.

$$\begin{aligned} \text{We have,} \quad & \vec{r}=3 \hat{i}-2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}+\lambda(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\ \text{and} \quad & \overrightarrow{\mathbf{r}}=(2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}})+\mu(6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\ \text{where,} \quad & \overrightarrow{\mathbf{a}_1}=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}} \quad \overrightarrow{\mathbf{b}_1}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \\ \text{and} \quad & \overrightarrow{\mathbf{a}_2}=2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}} \quad \overrightarrow{\mathbf{b}_2}=6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \end{aligned}$$

$$ \begin{aligned} &\text { If } \theta \text { is angle between the lines, then }\\ &\begin{aligned} \cos \theta & =\frac{\left|\overrightarrow{\mathbf{b}_1} \cdot \overrightarrow{\mathbf{b}_2}\right|}{\left|\overrightarrow{\mathbf{b}_1}\right| \cdot\left|\overrightarrow{\mathbf{b}_2}\right|} \\ & =\frac{|(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \cdot(6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})|}{|2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}||6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}|} \\ & =\frac{|12+3+4|}{\sqrt{9} \sqrt{49}}=\frac{19}{21} \\ \therefore\quad \theta & =\cos ^{-1} \frac{19}{21} \end{aligned} \end{aligned}$$

We know that, the cartesian equation of a line that passes through two points ( $x_1, y_1, z_1$ ) and $\left(x_2, y_2, z_2\right)$ is

$$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$$

Hence, the cartesian equation of line passes through $A(0,-1,-1)$ and $B(4,5,1)$ is

$$\begin{aligned} \frac{x-0}{4-0} & =\frac{y+1}{5+1}=\frac{z+1}{1+1} \\ \Rightarrow\quad \frac{x}{4} & =\frac{y+1}{6}=\frac{z+1}{2}\quad\text{.... (i)} \end{aligned}$$

and cartesian equation of the line passes through $C(3,9,4)$ and $D(-4,4,4)$ is

$$\begin{aligned} &\frac{x-3}{-4-3}=\frac{y-9}{4-9}=\frac{z-4}{4-4} \\ \Rightarrow\quad & \frac{x-3}{-7}=\frac{y-9}{-5}=\frac{z-4}{0}\quad\text{.... (ii)} \end{aligned}$$

If the lines intersect, then shortest distance between both of them should be zero.

$\therefore$ Shortest distance between the lines

$=\frac{\left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\end{array}\right|}{\sqrt{\left(b_1 c_2-b_2 c_1\right)^2+\left(c_1 a_2-c_2 a_1\right)^2+\left(a_1 b_2-a_2 b_1\right)^2}}$

$$\begin{aligned} & =\frac{\left|\begin{array}{ccc} 3-0 & 9+1 & 4+1 \\ 4 & 6 & 2 \\ -7 & -5 & 0 \end{array}\right|}{\sqrt{(6 \cdot 0+10)^2+(-14-0)^2+(-20+42)^2}} \\ & =\frac{\left|\begin{array}{ccc} 3 & 10 & 5 \\ 4 & 6 & 2 \\ -7 & -5 & 0 \end{array}\right|}{\sqrt{100+196+484}} \\ & =\frac{3(0+10)-10(14)+5(-20+42)}{\sqrt{780}} \\ & =\frac{30-140+110}{\sqrt{780}=0} \end{aligned}$$

So, the given lines intersect.