The equation of one of the straight lines which passes through the point

$y=m x+c$

$3=m+c$

$\sqrt{2}=\left|\frac{m-3 \sqrt{2}}{1+3 \sqrt{2} m}\right|$

$=6 m+\sqrt{2}=m-3 \sqrt{2}$

$=\sin =-4 \sqrt{2} \rightarrow m=\frac{-4 \sqrt{2}}{5}$

$==6 m-\sqrt{2}=m-3 \sqrt{2}$

$=7 m-2 \sqrt{2} \rightarrow m=\frac{2 \sqrt{2}}{7}$

According to options take $\mathrm{m}=\frac{-4 \sqrt{2}}{5}$

So $\mathrm{y}=\frac{-4 \sqrt{2} \mathrm{x}}{5}+\frac{3+4 \sqrt{2}}{5}$

$4 \sqrt{2} x+5 y-(15+4 \sqrt{2})=0$