Let the coordinates of moving point $P$ be $(x, y)$.

Given that, the sum of distances of this point in a plane from the axes is 1 .

$$\begin{array}{lr} \therefore & |x|+|y|=1 \\ \Rightarrow & \pm x \pm y=1 \\ \Rightarrow & x+y=1 \\ \Rightarrow & -x-y=1 \\ \Rightarrow & -x+y=1 \\ \Rightarrow & x-y=1 \end{array}$$

So, these equations give us locus of the point which is a square.

$$\begin{aligned} \text{Given equation of lines are} \quad & y-\sqrt{3} x=2 \quad [\because x\ge0]\\ \text{and}\quad & y+\sqrt{3} x=2 \quad [\because x\le0] \end{aligned}$$

$\because\quad y=\sqrt{3} x+2\quad\text{.... (i)}$

$$\begin{aligned} \text{and}\quad y & =-\sqrt{3} x+2 \\ \Rightarrow \quad \sqrt{3} x+2 & =-\sqrt{3} x+2 \\ \Rightarrow \quad 2 \sqrt{3} x & =0 \Rightarrow x=0 \end{aligned}$$

On putting $x=0$ in Eq. (i), we get

So, the point of intersection of line (i) and (ii) is ( 0,2 ).

Here, $OC=2$

In $\Delta DEC,$ $C D =\cos 30 \Upsilon $

$$\begin{aligned} \frac{O C}{C D} & =2 \\ C D & =5 \cos 30 \Upsilon \\ & =5 \cdot \frac{\sqrt{3}}{2} \\ O D & =O C+C D=2+5 \frac{\sqrt{3}}{2} \end{aligned}$$

So, the coordinates of the foot of perpendiculars are $\left(0,2+\frac{5 \sqrt{3}}{2}\right)$.

Given equation of line is,

$$\frac{x}{a}+\frac{y}{b}=1\quad \text{.... (i)}$$

Perpendicular length from the origin on the line (i) is given by $p$

$$\begin{aligned} \text{i.e.,}\quad & p=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}=\frac{a b}{\sqrt{a^2+b^2}} \\ \therefore \quad & p^2=\frac{a^2 b^2}{a^2+b^2} \end{aligned}$$

Given that, $a^2, p^2$ and $b^2$ are in AP.

$$ \begin{array}{lr} \therefore & 2 p^2=a^2+b^2 \\ \Rightarrow & \frac{2 a^2 b^2}{a^2+b^2}=a^2+b^2 \\ \Rightarrow & 2 a^2 b^2=\left(a^2+b^2\right)^2 \\ \Rightarrow & 2 a^2+b^2=a^4+b^4+2 a^2 b^2 \\ \Rightarrow & a^4+b^4=0 \end{array}$$