I think you meant:
adx/((b-c)yz)=bdy/((c-a)zx)=cdz/((a-b)xy).
If we take the first pair, we can eliminate z from the denominators:
adx/((b-c)y)=bdy/((c-a)x), now separate the variables:
a(c-a)xdx=b(b-c)ydy.
Assuming a, b, c are constants, integrate:
a(c-a)x²/2=b(b-c)y²/2+constant, which can be written:
a(c-a)x²=b(b-c)y²+C₁, where C₁ is a constant.
Similarly we can take the second pair and eliminate x, separate variables y and z and arrive at:
b(a-b)ydy=c(c-a)zdz, integrate:
b(a-b)y²=c(c-a)z²+C₂.
Finally, take the first and last to make a pair in which y can be eliminated. After integration, we have:
a(a-b)x²=c(b-c)z²+C₃.
By suitable manipulation (adding or subtracting the equations) and with the constants combined into one:
ax²(b+c-2a)+by²(a+c-2b)+cz²(a+b-2c)=C.