ZetaFlux has already observed that a cannot be even (he said it was easy to see, but it took me a while to see that it's because x and y and z must all be odd by the gcd requirement, and xy + xz + yz would then be odd)
let a = 2b+1
x = (2b+1)^2 * c^2 * (18b+10)  3*(2b+1)*c*(b+1) + b
y = (2b+1)^2 * c^2 * (54b+30)  (2b+1)*c*(9*b+1) + 1
z = (2b+1)^2 * c^2 * (27b+15) + (2b+1)*c*(9*b+7) + 1
Gives an infinite family of solutions, and the gcd requirement is met because
x = b mod (2b+1)
y = 1 mod (2b+1)
z = 1 mod (2b+1)
Thus there is an infinite family of solutions for all odd a.
