Let $z=(x-i y)(3+5 i)$

$z=3 x+5 x i-3 y i-5 y i^{2}=3 x+5 x i-3 y i+5 y=(3 x+5 y)+i(5 x-3 y)$

$\therefore \bar{z}=(3 x+5 y)-i(5 x-3 y)$

It is given that, $\bar{z}=-6-24 i$

$\therefore(3 x+5 y)-i(5 x-3 y)=-6-24 i$

Equating real and imaginary parts, we obtain

$3 x+5 y=-6$…..$(i)$

$5 x-3 y=24$….$(ii)$

Multiplying equation $(i)$ by $3$ and equation $(ii)$ by $5$ and then adding them, we obtain

$9 x+15 y=-18$

${25 x-15 y=120}$

$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

${34 x=102}$

$\therefore x=\frac{102}{34}=3$

Putting the value of $x$ in equation $(i),$ we obtain

$3(3)+5 y=-6$

$\Rightarrow 5 y=-6-9=-15$

$\Rightarrow y=-3$

Thus, the values of $x$ and $y$ are $3 $ and $-3$ respectively.