1. **State the problem:** We need to find the sum of the first 12 terms of a geometric series where the first term $a_1 = -3$ and the second term $a_2 = 6$. 2. **Find the common ratio $r$:** The common ratio is given by $r = \frac{a_2}{a_1} = \frac{6}{-3} = -2$. 3. **Formula for the sum of the first $n$ terms of a geometric series:** $$ S_n = a_1 \frac{1 - r^n}{1 - r} $$ 4. **Substitute the known values:** $$ S_{12} = -3 \frac{1 - (-2)^{12}}{1 - (-2)} $$ 5. **Calculate $(-2)^{12}$:** Since 12 is even, $(-2)^{12} = 2^{12} = 4096$. 6. **Simplify the numerator:** $$ 1 - 4096 = -4095 $$ 7. **Simplify the denominator:** $$ 1 - (-2) = 1 + 2 = 3 $$ 8. **Calculate the sum:** $$ S_{12} = -3 \times \frac{-4095}{3} $$ 9. **Cancel common factors:** $$ S_{12} = \cancel{-3} \times \frac{-4095}{\cancel{3}} = -1 \times (-4095) = 4095 $$ 10. **Final answer:** The sum of the first 12 terms is $4095$. **Answer choice:** c. 4095