Let intercept of a line are $(h, k)$.

The coordinates of $A$ and $B$ are $(h, 0)$ and $(0, k)$ respectively.

$$\begin{aligned} & -5=\frac{1 \times 0+2 \times h}{1+2} \\ \therefore \quad & -5=\frac{2 h}{3} \Rightarrow+h=-\frac{15}{2} \\ & \text { and } \\ & 4=\frac{1 \cdot k+0 \cdot 2}{1+2} \\ & k=12 \\ & A=\left(-\frac{15}{2}, 0\right) \text { and } B=(0,12) \end{aligned}$$

Hence, the equation of a line $A B$ is

$$\begin{aligned} & y-0 =\frac{12-0}{0+15 / 2}\left(x+\frac{15}{2}\right) \\ \Rightarrow \quad & y =\frac{12 \cdot 2}{15}\left(x+\frac{15}{2}\right) \\ \Rightarrow \quad & 5 y =8 x+60 \Rightarrow 8 x-5 y+60=0 \end{aligned}$$

$$\begin{aligned} &\begin{aligned} & \text { Given that, } \quad O C=P=4 \text { units } \\ & \angle B A X=120 \Upsilon\\ & \text { Let } \quad \angle C O A=\alpha, \angle O C A=90 \Upsilon \\ & \because \quad \angle B A X=\angle C O A+\angle O C A \quad \text { [exterior angle property] }\\ & \Rightarrow \quad 120 \Upsilon=\alpha+90 \Upsilon \\ & \therefore \quad \alpha=30 \Upsilon \end{aligned}\\ &\text { Now, the equation of required line is }\\ &\begin{array}{l} & x \cos 30 \Upsilon+y \sin 30 \Upsilon =4 \\ \Rightarrow & x \cdot \frac{\sqrt{3}}{2}+y \cdot \frac{1}{2} =4 \\ \Rightarrow & \sqrt{3} x+y =8 \end{array} \end{aligned}$$

Let slope of line $A C$ be $m$ and slope of line $B C$ is $\frac{-3}{4}$ and let angle between line $A C$ and $B C$ be $\theta$.

$$\begin{aligned} & \therefore \quad \tan \theta=\left|\frac{m+\frac{3}{4}}{1-\frac{3 m}{4}}\right| \Rightarrow \tan 45 \Upsilon= \pm\left[\frac{m+\frac{3}{4}}{1-\frac{3 m}{4}}\right] \\ & \text { Taking positive sign, } \\ & 1=\frac{m+\frac{3}{4}}{1-\frac{3 m}{4}} \\ & \Rightarrow \quad m+\frac{3}{4}=1-\frac{3 m}{4} \\ & \Rightarrow \quad m+\frac{3 m}{4}=1-\frac{3}{4} \\ & \Rightarrow \quad \frac{7 m}{4}=\frac{1}{4} \Rightarrow m=\frac{1}{7} \end{aligned}$$

Taking negative sign,

$$\begin{array}{rlrl} & 1 =-\left(\frac{m+\frac{3}{4}}{1-\frac{3 m}{4}}\right) \Rightarrow 1-\frac{3 m}{4}=-m-\frac{3}{4} \\ \Rightarrow \quad & m-\frac{3 m}{4} & =-1-\frac{3}{4} \\ \Rightarrow \quad & \frac{m}{4} & =\frac{-7}{4} \Rightarrow m=-7 \end{array}$$

$\therefore$ Equation of side $A C$ having slope $\left(\frac{1}{7}\right)$ is

$$\begin{array}{rlrl} & y-2 =\frac{1}{7}(x-2) \\ \Rightarrow \quad & 7 y-14 =x-2 \\ \Rightarrow \quad & x-7 y+12 =0 \end{array}$$

and equation of side $A B$ having slope (-7) is

$$\begin{aligned} & y-2 =-7(x-2) \\ \Rightarrow \quad & y-2 =-7 x+14 \\ \Rightarrow \quad & 7 x+y-16 =0 \end{aligned}$$

Given that, equilateral $\triangle A B C$ having equation of base is $x+y=2$.

$$\begin{aligned} & \text { In } \triangle A B D, \quad \sin 60 \gamma=\frac{A D}{A B} \\ & \Rightarrow \quad A D=A B \sin 60 \Upsilon=A B \frac{\sqrt{3}}{2} \\ \because \quad & A D=A B \frac{\sqrt{3}}{2}\quad \text{... (i)} \end{aligned}$$

Now, the length of perpendicular from $(2,-1)$ to the line $x+y=2$ is given by

$$\begin{aligned} & A D=\left|\frac{2+(-1)-2}{\sqrt{1^2+1^2}}\right|=\frac{1}{\sqrt{2}} \\ & \frac{1}{\sqrt{2}}=A B \frac{\sqrt{3}}{2} \\ \text{From Eq. (i),}\quad & A B=\sqrt{\frac{2}{3}} \end{aligned}$$

Let slope of the line be $m$ and the coordinates of fixed point $P$ are $\left(x_1, y_1\right)$.

$\therefore \quad$ Equation of line is $y-y_1=m\left(x-x_1\right)\quad \text{.... (i)}$

Since, the given points are $A(2,0), B(0,2)$ and $C(1,1)$.

Now, perpendicular distance from $A$, is

$$\frac{0-y_1-m\left(2-x_1\right)}{\sqrt{1+m^2}}$$

Perpendicular distance from $B$, is

$$\frac{2-y_1-m\left(0-x_1\right)}{\sqrt{1+m^2}}$$

Perpendicular distance from $C$, is

$$\frac{1-y_1-m\left(1-x_1\right)}{\sqrt{1+m^2}}$$

Now, $\quad \frac{-y_1-2 m+m x_1+2-y_1+m x_1+1-y_1-m+m x_1}{\sqrt{1+m^2}}=0$

$$\begin{array}{ll} \Rightarrow & -3 y_1-3 m+3 m x_1+3=0 \\ \Rightarrow & -y_1-m+m x_1+1=0 \end{array}$$

Since, $(1,1)$ lies on this line. So, the point $P$ is $(1,1)$.