If $9\left(a^2+b^2\right)+c^2+20=12(a+2 b)$, then the value of $\sqrt{6 a+9 b+2 c}$ is :

4

We know that,

( a - b )² = a² + b² - 2ab

( a + b )² = a² + b² + 2ab

9(a² + b² ) + c² + 20 = 12(a + 2b)

= 9a² + 9b² + c² + 20 = 12a + 24b

= 9a² - 12a + 9b² - 24b + c² + 20 = 0

= 9a² - 12a + 4 + 9b² - 24b + 16 + c² = 0

= (3a - 2)² + (3b - 4)² + c² = 0

3a – 2 = 0

a = \(\frac{2}{3}\)

3b  - 4 = 0

b = \(\frac{4}{3}\)

c = 0

$\sqrt{6 a+9 b+2 c}$ = $\sqrt{6 (\frac{2}{3}) +9 (\frac{4}{3})+2 (0)}$

$\sqrt{6 a+9 b+2 c}$ = \(\sqrt {16}\) = 4