Let the moving point be $P(x,y)$.

Its distances from $(2,1)$ and $(1,3)$ are in the ratio $5:4$. So, $\frac{\sqrt{(x-2)^2+(y-1)^2}}{\sqrt{(x-1)^2+(y-3)^2}}=\frac{5}{4}$

Squaring both sides: $\frac{(x-2)^2+(y-1)^2}{(x-1)^2+(y-3)^2}=\frac{25}{16}$

Hence, $16\big[(x-2)^2+(y-1)^2\big]=25\big[(x-1)^2+(y-3)^2\big]$

Expand both sides.

Left side: $16\left[(x^2-4x+4)+(y^2-2y+1)\right]$ $=16(x^2+y^2-4x-2y+5)$ $=16x^2+16y^2-64x-32y+80$

Right side: $25\left[(x^2-2x+1)+(y^2-6y+9)\right]$ $=25(x^2+y^2-2x-6y+10)$ $=25x^2+25y^2-50x-150y+250$

Bring all terms to one side: $16x^2+16y^2-64x-32y+80-25x^2-25y^2+50x+150y-250=0$

$-9x^2-9y^2-14x+118y-170=0$

Multiplying by $-1$: $9x^2+9y^2+14x-118y+170=0$

Compare with $ax^2+by^2+cxy+dx+ey+170=0$

So, $a=9,\quad b=9,\quad c=0,\quad d=14,\quad e=-118$

Compute the required value: $a^2+2b+3c+4d+e$ $=9^2+2(9)+3(0)+4(14)-118$ $=81+18+0+56-118$ $=37$

Therefore, the correct option is: $\boxed{37}$ i.e. Option A.