Given that 2a-3b + C = 0, 3a-2b-6c = 0, find the value of a & sup2; + B & sup2; + C & sup2; / 2A & sup2; + B & sup2; - C & sup2

Given that 2a-3b + C = 0, 3a-2b-6c = 0, find the value of a & sup2; + B & sup2; + C & sup2; / 2A & sup2; + B & sup2; - C & sup2

2a-3b+c=0.(1)
3a-2b-6c=0.(2)
Formula (1) × 3: 6a-9b + 3C = 0. (3)
Formula (2) × 2: 6a-4b-12c = 0. (4)
(4) (3) get: 5b-15c = 0, that is: B = 3C
Substituting B = 3C into formula (1), we get 2a-9c + C = 0, that is, a = 4C
Substituting a = 4C and B = 3C into a & sup2; + B & sup2; + C & sup2; / 2A & sup2; + B & sup2; - C & sup2;, we get the following results:
The original formula = (16C & sup2; + 9C & sup2; + C & sup2;) / (2 × 16C & sup2; + 9C & sup2; - C & sup2;)
=26c²/40c²
=26/40
=13/20