Since, normal to the plane is equally inclined to the coordinate axis.

Therefore, $$\cos \alpha=\cos \beta=\cos \gamma=\frac{1}{\sqrt{3}}$$

So, the normal is $\overrightarrow{\mathbf{N}}=\frac{1}{\sqrt{3}} \hat{\mathbf{i}}+\frac{1}{\sqrt{3}} \hat{\mathbf{j}}+\frac{1}{\sqrt{3}} \hat{\mathbf{k}}$ and plane is at a distance of $3 \sqrt{3}$ units from origin.

The equation of plane is $$\overrightarrow{\mathbf{r}} \cdot \hat{\mathbf{N}}=3 \sqrt{3}\quad \left[\because \hat{\mathbf{N}}=\frac{\overrightarrow{\mathbf{N}}}{|\mathbf{N}|}\right]$$

[since, vector equation of the plane at a distance $p$ from the origin is $\overrightarrow{\mathbf{r}} \cdot \hat{\mathbf{N}}=p$]

$$\begin{array}{rlrl} \Rightarrow & (x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot \frac{\left(\frac{1}{\sqrt{3}} \hat{\mathbf{i}}+\frac{1}{\sqrt{3}} \hat{\mathbf{j}}+\frac{1}{\sqrt{3}} \hat{\mathbf{k}}\right)}{1} =3 \sqrt{3} \\ \Rightarrow & \frac{x}{\sqrt{3}}+\frac{y}{\sqrt{3}}+\frac{z}{\sqrt{3}} =3 \sqrt{3} \\ \therefore & x+y+z =3 \sqrt{3} \cdot \sqrt{3}=9 \end{array}$$

So, the required equation of plane is $x+y+z=9$.

Since, the line drawn from the point $(-2,-1,-3)$ meets a plane at right angle at the point $(1,-3,3)$. So, the plane passes through the point $(1,-3,3)$ and normal to plane is $(-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}})$.

$$\begin{aligned} &\begin{array}{ll} \Rightarrow & \overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ \text { and } & \overrightarrow{\mathbf{N}}=-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}} \end{array}\\ &\text { So, the equation of required plane is }(\vec{r}-\overrightarrow{\mathbf{a}}) \cdot \overrightarrow{\mathbf{N}}=0\\ &\begin{aligned} \Rightarrow & {[(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}})-(\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})] \cdot(-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}) } =0 \\ \Rightarrow & {[(x-1) \hat{\mathbf{i}}+(y+3) \hat{\mathbf{j}}+(z-3) \hat{\mathbf{k}}] \cdot(-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}) } =0 \\ \Rightarrow & -3 x+3+2 y+6-6 z+18 =0 \\ \Rightarrow & -3 x+2 y-6 z =-27 \\ \therefore & -2 y+6 z-27 =0 \end{aligned} \end{aligned}$$

We know that, the equation of a plane passing through three non-collinear points $\left(x_1, y_1, z_1\right)$, $\left(x_2, y_2, z_2\right)$ and $\left(x_3, y_3, z_3\right)$ is

$$\left|\begin{array}{ccc} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{array}\right|=0$$

$$\begin{array}{ll} \Rightarrow & \left|\begin{array}{ccc} x-2 & y-1 & z-0 \\ 3-2 & -2-1 & -2-0 \\ 3-2 & 1-1 & 7-0 \end{array}\right|=0 \\ \Rightarrow & \left|\begin{array}{ccc} x-2 & y-1 & z \\ 1 & -3 & -2 \\ 1 & 0 & 7 \end{array}\right|=0 \end{array}$$

$$\begin{aligned} &\begin{array}{lrl} \Rightarrow & (x-2)(-21+0)-(y-1)(7+2)+z(3) & =0 \\ \Rightarrow & -21 x+42-9 y+9+3 z & =0 \\ \Rightarrow & -21 x-9 y+3 z & =-51 \\ \therefore & 7 x+3 y-z & =17 \end{array}\\ &\text { So, the required equation of plane is } 7 x+3 y-z=17 \text {. } \end{aligned}$$

Given equation of the line is $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}=\lambda\quad\text{.... (i)}$

So, DR's of the line are 2, 1, 1 and DC's of the given line are $\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{16}}$.

Also, the required lines make angle $\frac{\pi}{3}$ with the given line.

$$\begin{array}{lrl} \text { From Eq. (i), } & x & =(2 \lambda+3), y=(\lambda+3) \text { and } z=\lambda \\ \because & \cos \theta & =\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}} \\ \therefore & \cos \frac{\pi}{3} & =\frac{(4 \lambda+6)+(\lambda+3)+(\lambda)}{\sqrt{6} \sqrt{(2 \lambda+3)^2+(\lambda+3)^2+\lambda^2}} \end{array}$$

$$\begin{array}{ll} \Rightarrow & \frac{1}{2}=\frac{6 \lambda+9}{\sqrt{6} \sqrt{\left(4 \lambda^2+9+12 \lambda+\lambda^2+9+6 \lambda+\lambda^2\right)}} \\ \Rightarrow & \frac{\sqrt{6}}{2}=\frac{6 \lambda+9}{\sqrt{6 \lambda^2+18 \lambda+18}} \end{array}$$

$$\begin{array}{lrl} \Rightarrow & 6 \sqrt{\left(\lambda^2+3 \lambda+3\right)} & =2(6 \lambda+9) \\ \Rightarrow & 36\left(\lambda^2+3 \lambda+3\right) & =36\left(4 \lambda^2+9+12 \lambda\right) \\ \Rightarrow & \lambda^2+3 \lambda+3 & =4 \lambda^2+9+12 \lambda \\ \Rightarrow & 3 \lambda^2+9 \lambda+6 & =0 \\ \Rightarrow & \lambda^2+3 \lambda+2 & =0 \\ \Rightarrow & \lambda(\lambda+2)+1(\lambda+2) & =0 \\ \Rightarrow & (\lambda+1)(\lambda+2) & =0 \end{array}$$

$\therefore \quad \lambda=-1,-2$

So, the DC's are $1,2,-1$ and $-1,1,-2$.

Also, both the required lines passes through origin.

So, the equations of required lines are $\frac{x}{1}=\frac{y}{2}=\frac{z}{-1}$ and $\frac{x}{-1}=\frac{y}{1}=\frac{z}{-2}$.

$$\begin{aligned} &\text { Eliminating } n \text { from both the equations, we have }\\ &\begin{aligned} & l^2+m^2-(l-m)^2=0 \\ & \Rightarrow \quad l^2+m^2-l^2-m^2+2 m l=0 \quad \Rightarrow \quad 2 l m=0 \end{aligned} \end{aligned}$$

$$ \begin{aligned} & \Rightarrow \quad l m=0 \quad \Rightarrow \quad(-m-n) m=0 \quad[\because l=-m-n] \\ & \Rightarrow \quad(m+n) m=0 \\ & \Rightarrow \quad m=-n \Rightarrow m=0 \\ & \Rightarrow \quad l=0, l=-n \end{aligned}$$

Thus, Dr's two lines are proportional to $0,-n, n$ and $-n, 0, n$ i.e., $0,-1,1$ and $-1,0,1$.

So, the vector parallel to these given lines are $\overrightarrow{\mathbf{a}}=-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}=-\hat{\mathbf{i}}+\hat{\mathbf{k}}$

$$ \begin{aligned} &\text { Now, }\\ &\begin{aligned} \cos \theta & =\frac{\vec{a} \vec{b}}{|\vec{a}||\vec{b}|}=\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \Rightarrow \cos \theta=\frac{1}{2} \\ \therefore\quad \theta & =\frac{\pi}{3} \quad\left[\because \cos \frac{\pi}{3}=\frac{1}{2}\right] \end{aligned} \end{aligned}$$