Let $\gcd(a,b) = z, \gcd(a,c) = y, \gcd(b,c) = x$ So, $a = yza_1, b = xzb_1, c= xyc_1$ for some positive integer $a_1,b_1,c_1$ such that $\gcd(ya_1,xb_1) = \gcd(zb_1, yc_1) = \gcd(za_1,xc_1) =1$ By given condition, we have $$a_1b_1c_1 \mid x^2b_1^2c_1^2 + y^2a_1^2c_1^2 + z^2 a_1^2b_1^2$$Which implies $$a_1 \mid x^2b_1^2c_1^2$$but $\gcd(ya_1,xb_1) = \gcd(za_1,xc_1) = \gcd(a_1,x^2b_1^2c_1^2) =1$, thus we have $a_1 = 1$, which follows that $a_1 = b_1 = c_1 = 1$ So, we have $$a = yz, b = xz, c= xy \rightarrow abc = (xyz)^2$$as desired $\blacksquare$