1. **State the problem:** Given the arithmetic sequence terms $a_2, a_6, a_{10}, a_{14}$ and the equation $$3(a_2 + a_6) + 2(a_6 + a_{10} + a_{14}) = 24,$$ find the sum of the first 13 terms, $S_{13}$. 2. **Recall the formula for the $n$th term of an arithmetic sequence:** $$a_n = a_1 + (n-1)d,$$ where $a_1$ is the first term and $d$ is the common difference. 3. **Express the given terms in terms of $a_1$ and $d$:** $$a_2 = a_1 + d,$$ $$a_6 = a_1 + 5d,$$ $$a_{10} = a_1 + 9d,$$ $$a_{14} = a_1 + 13d.$$ 4. **Substitute into the given equation:** $$3(a_2 + a_6) + 2(a_6 + a_{10} + a_{14}) = 24,$$ $$3[(a_1 + d) + (a_1 + 5d)] + 2[(a_1 + 5d) + (a_1 + 9d) + (a_1 + 13d)] = 24.$$ 5. **Simplify inside the brackets:** $$3(2a_1 + 6d) + 2(3a_1 + 27d) = 24,$$ $$6a_1 + 18d + 6a_1 + 54d = 24,$$ $$12a_1 + 72d = 24.$$ 6. **Divide both sides by 12:** $$\cancel{12}a_1 + \cancel{12}6d = \cancel{12}2,$$ $$a_1 + 6d = 2.$$ 7. **Recall the formula for the sum of the first $n$ terms:** $$S_n = \frac{n}{2}[2a_1 + (n-1)d].$$ 8. **Find $S_{13}$:** $$S_{13} = \frac{13}{2}[2a_1 + 12d].$$ 9. **Rewrite $2a_1 + 12d$ as $2(a_1 + 6d)$:** $$S_{13} = \frac{13}{2} \times 2(a_1 + 6d) = 13(a_1 + 6d).$$ 10. **Substitute $a_1 + 6d = 2$ from step 6:** $$S_{13} = 13 \times 2 = 26.$$ **Final answer:** $S_{13} = 26$. The correct choice is **D. 26**.