Given points are $A(3,4), B(-3,6), C(-5,1), D(x,y)$.

Here area of $ABC=$Area of $ACD$.

Area of $ABC=\frac{1}{2}\left|\sum x_1\left(y_2-y_3\right)\right|$

$=\frac{1}{2}\left|3\left(6-1\right)+\left(-3\right)\left(1-4\right)+\left(-5\right)\left(4-6\right)\right|$

$=\frac{1}{2}\left|3\left(5\right)+\left(-3\right)\left(-3\right)+\left(-5\right)\left(-2\right)\right|$

$=\frac{1}{2}\left|15+9+10\right|$

$=17 ...\left(1\right)$

Area of $ACD=\frac{1}{2}\left|\sum x_1\left(y_2-y_3\right)\right|$

$=\frac{1}{2}\left|3\left(1-y\right)+\left(-5\right)\left(y-4\right)+x\left(4-1\right)\right|$

$=\frac{1}{2}\left|3-3y-5y+20+3x\right|$

$=\frac{1}{2}\left|+3x-8y+23\right| ...\left(2\right)$

From equations $\left(1\right)$ and $\left(2\right)$,

$\pm (+3x-8y+23)=34$

$\Rightarrow 3x-8y+23=\pm 34$

$\Rightarrow 3x-8y+23=34$ or $3x-8y+23=-34$

$\Rightarrow 3x-8y-11=0$ or $3x-8y+57=0$

$\therefore$ The locus of $D$ is $\left(3x-8y-11\right)\left(3x-8y+57\right)=0$.