3. What is $3 S^{2}$ if
\[
S=\sum_{n=0}^{\infty}\left(\frac{12^{n}}{21^{n+1}}\right) ?
\]
$\frac{1}{27}$
$\frac{1}{9}$
$\frac{1}{3}$
$\frac{1}{81}$
Diverges

Step 1 ：Given the geometric series \(S=\sum_{n=0}^{\infty}\left(\frac{12^{n}}{21^{n+1}}\right)\), we need to find the value of \(3 S^{2}\).

Step 2 ：The sum of a geometric series is given by the formula \(S = \frac{a}{1-r}\), where \(a\) is the first term and \(r\) is the common ratio.

Step 3 ：In this case, the first term \(a\) is \(\frac{12^{0}}{21^{1}} = \frac{1}{21}\) and the common ratio \(r\) is \(\frac{12}{21}\).

Step 4 ：Substituting these values into the formula, we find that \(S = \frac{1/21}{1-(12/21)} = 0.11111111111111109\).

Step 5 ：Finally, we calculate \(3S^2 = 3*(0.11111111111111109)^2 = 0.03703703703703702\).

Step 6 ：Thus, the value of \(3 S^{2}\) is \(\boxed{\frac{1}{27}}\).