The number of elements in the set $$S = \left\{(r, k) : k \in \mathbb{Z} \text{ and } {}^{36}C_{r+1} = \frac{6 \cdot {}^{35}C_r}{k^2 - 3}\right\}$$, is :

We start with the given equation $${}^{36}C_{r+1} = \frac{6 \cdot {}^{35}C_r}{k^2 - 3}$$ and compute the ratio:

$$\frac{{}^{36}C_{r+1}}{{}^{35}C_r} = \frac{36!}{(r+1)!(35-r)!} \cdot \frac{r!(35-r)!}{35!} = \frac{36}{r+1}$$

So the equation becomes $$\frac{36}{r+1} = \frac{6}{k^2 - 3}$$, which gives:

$$k^2 - 3 = \frac{r+1}{6} \implies k^2 = \frac{r + 19}{6}$$

For $$k \in \mathbb{Z}$$, we need $$r + 19$$ to be divisible by 6 and $$\frac{r+19}{6}$$ to be a perfect square. Also, $$0 \leq r \leq 35$$.

From $$r + 19 \equiv 0 \pmod{6}$$, we get $$r \equiv 5 \pmod{6}$$. The possible values are $$r = 5, 11, 17, 23, 29, 35$$.

Checking each:

$$r = 5$$: $$k^2 = \frac{24}{6} = 4$$, so $$k = \pm 2$$ (valid)

$$r = 11$$: $$k^2 = \frac{30}{6} = 5$$ (not a perfect square)

$$r = 17$$: $$k^2 = \frac{36}{6} = 6$$ (not a perfect square)

$$r = 23$$: $$k^2 = \frac{42}{6} = 7$$ (not a perfect square)

$$r = 29$$: $$k^2 = \frac{48}{6} = 8$$ (not a perfect square)

$$r = 35$$: $$k^2 = \frac{54}{6} = 9$$, so $$k = \pm 3$$ (valid)

The set $$S = \{(5, 2), (5, -2), (35, 3), (35, -3)\}$$, which has 4 elements.

Hence, the correct answer is Option 2.