-x^2 - (3*4-2*7)z^2 + (3*54 - 2*74) = 0

-x^2 + 2z^2 + 14 = 0

Eliminate z by subtracting 4x the second from 7x the first equation:

(3*7-5*4)x^2 - (7*2 - 4*3)y^2 + (7*54 - 4*74) = 0

x^2 - 2y^2 + 82 = 0

Now add these two:

2z^2 - 2y^2 + 96 = 0

(y^2 - z^2) = 48

(y+z)(y-z) = 48

Since y and z are integers there are a finite number of these. The possible factors are: (48,1), (24,2), (16,3), (12,4), and (8,6)

But (48,1) and (16,3) are impossible because they'd give non-integer values of y -- If the two factors are a and b then a+b = 2y which is even, so a and b must have the same parity. Then there are only three options:

(24,2) => (y,z) = (13,11)

(12,4) => (y,z) = (8,4)

(8,6) => (y,z) = (7,1)

Now we just need to see what x would be for these cases, which we can do by substituting y and z in either equation.

x^2 = (2y^2 + 4z^2 - 54) / 3 so

(y,z) = (13,11) => x^2 = (2*169 + 4*121 - 54) / 3 = 768/3 = 256 = 16^2 so (16,13,11) is a solution

(y,z) = (8,4) => x^2 = (2*64 + 4*16 - 54) / 3 = 138/3 = 46 which isn't a perfect square, so there's no solution here

(y,z) = (7,1) => x^2 = (2*49 + 4*1 - 54) / 3 = 48/3 = 16 = 4^2 so (4,7,1) is a solution.

That exhausts all the ways to factor 48, so there are no others.