The value of $$\lim_{n \to \infty} 6 \tan\left\{\sum_{r=1}^{n} \tan^{-1}\left(\frac{1}{r^2 + 3r + 3}\right)\right\}$$ is equal to

We need to evaluate $$\displaystyle\lim_{n \to \infty} 6\tan\left\{\sum_{r=1}^{n} \tan^{-1}\left(\frac{1}{r^2 + 3r + 3}\right)\right\}$$.

Simplify the general term using a telescoping identity:

Note that $$r^2 + 3r + 3 = 1 + (r + 1)(r + 2)$$.

Using the identity $$\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\!\left(\frac{A - B}{1 + AB}\right)$$, we observe:

$$\tan^{-1}(r + 2) - \tan^{-1}(r + 1) = \tan^{-1}\!\left(\frac{(r+2) - (r+1)}{1 + (r+1)(r+2)}\right) = \tan^{-1}\!\left(\frac{1}{1 + (r+1)(r+2)}\right)$$

$$= \tan^{-1}\!\left(\frac{1}{r^2 + 3r + 3}\right)$$

Evaluate the telescoping sum:

$$\sum_{r=1}^{n} \tan^{-1}\!\left(\frac{1}{r^2 + 3r + 3}\right) = \sum_{r=1}^{n} \left[\tan^{-1}(r + 2) - \tan^{-1}(r + 1)\right]$$

$$= \tan^{-1}(n + 2) - \tan^{-1}(2)$$

Take the limit as $$n \to \infty$$:

$$\lim_{n \to \infty} \left[\tan^{-1}(n + 2) - \tan^{-1}(2)\right] = \frac{\pi}{2} - \tan^{-1}(2)$$

Compute the final expression:

$$6\tan\!\left(\frac{\pi}{2} - \tan^{-1}(2)\right) = 6\cot\!\left(\tan^{-1}(2)\right) = 6 \times \frac{1}{2} = 3$$

The correct answer is Option C: $$3$$.