Solve the equations AX = by = CZ = 1 / x + 1 / y + 1 / Z, where a > 0, b > 0, C > 0

Solve the equations AX = by = CZ = 1 / x + 1 / y + 1 / Z, where a > 0, b > 0, C > 0

Y = ax / B, z = ax / C, substituting into the equation AX = 1 / x + 1 / y + 1 / Z, we can get the solution of the equation AX = (a + B + C) / ax, and get | x | = radical (a + B + C) / A

AX+BY+CZ=2 …… A study on the system of linear equations with three variables in the sixth grade
If the solution of the system {ax + by + CZ = 2 BX + CY + AZ = 2 CX + ay + BZ = 2 is {x = 1 y = - 2 z = 3, find the values of a, B and C

A-2b + 3C = 2, b-2c + 3A = 2c-2a + 3B = 2, a-2b + 3C = 2, a-3a + b-2c = 2, b-2a + 3B + C = 2, b-4c = 2, b-4c = 2, b-2c = 2, b-2c = 2, C = 2, C = 2, C = 2, C = 1, a = 1, a = 1, so a = 1, B = 1, B = 1

Solving the equations ax + by = 3C ① ax + CZ = 3b ② by + CZ = 3A ③

① + 2 + 3
2(ax+by+cz)=3a+3b+3c
Ax + by + CZ = (3a + 3B + 3C) / 2 denotes ④
④ (1) CZ = (3a + 3b-3c) / 2, z = (3a + 3b-3c) / (2C)
④ (2) by = (3a-3b + 3C) / 2, y = (3a-3b + 3C) / (2b)
④ (3) AX = (- 3A + 3B + 3C) / 2, x = (- 3A + 3B + 3C) / (2a)