14. Given
$$
\frac{1}{\log _{2} a}+\frac{1}{\log _{3} a}+\frac{1}{\log _{4} a}=1 \text {. }
$$

Then the value of $a$ is ( ).
(A) 9
(B) 12
(C) 18
(D) 24
(E) 36
14. D.

Notice,
$$
\begin{aligned}
& \frac{1}{\log _{2} a}+\frac{1}{\log _{3} a}+\frac{1}{\log _{4} a} \\
= & \log _{a} 2+\log _{a} 3+\log _{a} 4 \\
= & \log _{a}(2 \times 3 \times 4)=\log _{a} 24=1 .
\end{aligned}
$$

Thus, $a=24$.