The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12. If $$\mu$$ and $$\sigma^2$$ denote the mean and variance of the correct observations respectively, then $$15(\mu + \mu^2 + \sigma^2)$$ is equal to _____.

Original: $$n = 15$$, $$\bar{x} = 12$$, $$\sigma = 3$$, $$\sigma^2 = 9$$.

$$\sum x_i = 15 \times 12 = 180$$. $$\sum x_i^2 = n(\sigma^2 + \bar{x}^2) = 15(9 + 144) = 15 \times 153 = 2295$$.

Corrected: Replace 10 with 12.

$$\sum x_i' = 180 - 10 + 12 = 182$$. $$\mu = 182/15$$.

$$\sum x_i'^2 = 2295 - 100 + 144 = 2339$$.

$$\sigma^2 = \frac{2339}{15} - \left(\frac{182}{15}\right)^2 = \frac{2339}{15} - \frac{33124}{225} = \frac{35085 - 33124}{225} = \frac{1961}{225}$$.

$$15(\mu + \mu^2 + \sigma^2) = 15\mu + 15\mu^2 + 15\sigma^2$$

$$= 182 + 15\left(\frac{182}{15}\right)^2 + 15 \times \frac{1961}{225}$$

$$= 182 + \frac{182^2}{15} + \frac{1961}{15}$$

$$= 182 + \frac{33124 + 1961}{15} = 182 + \frac{35085}{15} = 182 + 2339 = 2521$$

The answer is $$\boxed{2521}$$.