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

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,

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

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

Given equation of the line is

$$x \sin \theta+y \cos \theta=p\quad\text{... (i)}$$

Let the mid-point of $A B$ is $p(h, k)$.

So, the mid-point of $A B$ are $$\left(\frac{a}{2}, \frac{b}{2}\right)$$

Since, the point $(a, 0)$ lies on the line $(\mathrm{i})$, then

$$\begin{aligned} a \sin \theta+0 & =p \\ a \sin \theta & =p \Rightarrow a=\frac{p}{\sin \theta} \end{aligned}$$

and the point $(0, b)$ also lies on the line, then

$$\begin{aligned} & 0+b \cos \theta=p \\ & \Rightarrow \quad b \cos \theta=p \Rightarrow b=\frac{p}{\cos \theta} \\ & \text { Now, mid-point of } A B=\left(\frac{a}{2}, \frac{b}{2}\right) \text { or }\left(\frac{p}{2 \sin \theta}, \frac{p}{2 \cos \theta}\right) \\ & \because \quad \frac{p}{2 \sin \theta}=h \Rightarrow \sin \theta=\frac{p}{2 h} \\ & \text { and } \quad \frac{p}{2 \cos \theta}=k \Rightarrow \cos \theta=\frac{p}{2 k} \\ & \therefore \quad \sin ^2 \theta+\cos ^2 \theta=\frac{p^2}{4 h^2}+\frac{p^2}{4 k^2} \\ & \Rightarrow \quad 1=\frac{p^2}{4}\left(\frac{1}{h^2}+\frac{1}{k^2}\right) \end{aligned}$$

Locus of the mid-point is

$$4=p^2\left(\frac{1}{x^2}+\frac{1}{y^2}\right)$$

$$\Rightarrow \quad 4 x^2 y^2=p^2\left(x^2+y^2\right)$$