If \( \mu \) and \( \sigma^2 \) are mean and variance of a random variable \( X \) whose p.m.f. is given by
\[
P(X = x) = \binom{6}{x} \left( \frac{1}{4} \right)^x \left( \frac{3}{4} \right)^{6 - x}, \quad x = 0, 1, 2, 3, 4, 5, 6, \text{ then the value of } 2\mu + 12\sigma^2 =
\]