Given line is $$a x+b y+c=0\quad \text{.... (i)}$$

Since, $a, b$ and $c$ are in AP, then

$$\begin{aligned} b & =\frac{a+c}{2} \\ a-2 b+c & =0\quad \text{.... (ii)} \end{aligned}$$

On comparing Eqs.(i) and (ii), we get

$$x=1, y=2\quad \text{[using value of b in Eq. (i)]}$$

So, $(1,-2)$ lies on the line.

Let equation of line is

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

Since, this line passes through $(1,-2)$.

$$\begin{aligned} & \frac{1}{a}-\frac{2}{a}=1 \\ \Rightarrow \quad & 1-2=a \Rightarrow a=-1 \end{aligned}$$

$\therefore$ Required equation of the line is

$$\begin{aligned} -x-y & =1 \\ \Rightarrow \quad x+y+1 & =0 \end{aligned}$$

Since, the given point $P(3,2)$ and line is $x-2 y=3$

Slope of this line is $m_1=\frac{1}{2}$

Let the slope of the required line is $m$.

Then,

$$\begin{aligned} \tan \theta & =\left|\frac{m-\frac{1}{2}}{1+\frac{1}{2} m}\right| \\ 1 & = \pm\left(\frac{m-\frac{1}{2}}{1+\frac{m}{2}}\right) \quad [\because \tan45^\circ=1] \text{... (i)} \end{aligned}$$

Taking positive sign, $$1+\frac{m}{2}=m-\frac{1}{2}$$

$$\begin{aligned} &\begin{array}{ll} \Rightarrow & m-\frac{m}{2}=1+\frac{1}{2} \\ \Rightarrow & \frac{m}{2}=\frac{3}{2} \Rightarrow m=3 \end{array}\\ &\text { Taking negative sign, } \end{aligned}$$

$$\begin{array}{ll} 1 & =-\left(\frac{m-\frac{1}{2}}{1+\frac{m}{2}}\right) \\ \Rightarrow & 1+\frac{m}{2}=-m+\frac{1}{2} \\ \Rightarrow & m+\frac{m}{2}=\frac{1}{2}-1 \\ \Rightarrow & \frac{3 m}{2}=\frac{-1}{2} \Rightarrow m=\frac{-1}{3} \end{array}$$

$\therefore$ First equation of the line is

$$\begin{aligned} & y-2=3(x-3) \\ & \Rightarrow \quad 3 x-y-7=0 \end{aligned}$$

and second equation of the line is

$$\begin{aligned} & y-2=-\frac{1}{3}(x-3) \\ & \Rightarrow \quad 3 y-6=-x+3 \\ & \Rightarrow \quad x+3 y-9=0 \end{aligned}$$

$$\begin{array}{lc} \text { Given line is } & 3 x-4 y-8=0 \quad \text{... (i)}\\ \text { For point }(3,4), & 9-4 \cdot 4-8 \\ \Rightarrow & 9-16-8 \\ \Rightarrow & 9-24 \\ \Rightarrow & -15<0\quad \text{.... (i)} \end{array}$$

For point $(2,-6)$,

$$\begin{aligned} 6 & +24-8 \\ 22 & >0\quad \text{... (ii)} \end{aligned}$$

Since, the value are of opposite sign.

Hence, the points $(3,4)$ and $(2,-6)$ lies on opposite side to the line.

Let the coordinaters of the point are $(h, k)$,

$\therefore \quad$ Distance between $(3,-2)$ and $(h, k)$,

$$d_1^2=(3-h)^2+(-2-k)^2\quad \text{... (i)}$$

Now, distance of the point $(h, k)$ from the line $5 x-12 y=3$ is,

$$d_2=\left|\frac{5 h-12 k-3}{\sqrt{25+144}}\right|=\left|\frac{5 h-12 k-3}{13}\right|\quad \text{.... (ii)}$$

$$\begin{aligned} &\begin{aligned} & \text { Given that, } \quad d_1^2=d_2 \\ & \Rightarrow \quad(3-h)^2+(2+k)^2=\frac{5 h-12 k-3}{13} \\ & \Rightarrow \quad 9-6 h+h^2+4+4 k+k^2=\frac{5 h-12 k-3}{13} \\ & \Rightarrow \quad h^2+k^2-6 h+4 k+13=\frac{5 h-12 k-3}{13} \\ & \Rightarrow \quad 13 h^2+13 k^2-78 h+52 k+169=5 h-12 k-3 \\ & \Rightarrow \quad 13 h^2+13 k^2-83 h+64 k+172=0 \end{aligned}\\ &\therefore \text { Locus of this point is }\\ &13 x^2+13 y^2-83 x+64 y+172=0 \end{aligned}$$