If $$\left(3a-\frac{1}{b}\right)^{2}-8\left(3a-\frac{1}{b}\right)+16+\left(c+\frac{1}{b}-2a\right)\left(3a-\frac{1}{b}-4\right)=0$$, then

A

a + 3c + = 4

B

a + c + = 4

C

3a + c + = 4

D

2a + 3c + = 4