1. The problem asks for the 12th term of the harmonic sequence with terms 6, 4, 3, ... 2. A harmonic sequence is a sequence whose terms are the reciprocals of an arithmetic sequence. So first, find the arithmetic sequence formed by the reciprocals of the given terms: $$a_1 = \frac{1}{6}, a_2 = \frac{1}{4}, a_3 = \frac{1}{3}$$ 3. Find the common difference $d$ of this arithmetic sequence: $$d = a_2 - a_1 = \frac{1}{4} - \frac{1}{6} = \frac{3}{12} - \frac{2}{12} = \frac{1}{12}$$ 4. The $n$th term of an arithmetic sequence is given by: $$a_n = a_1 + (n-1)d$$ 5. Find the 12th term of the arithmetic sequence: $$a_{12} = \frac{1}{6} + (12-1) \times \frac{1}{12} = \frac{1}{6} + \frac{11}{12} = \frac{2}{12} + \frac{11}{12} = \frac{13}{12}$$ 6. The 12th term of the harmonic sequence is the reciprocal of $a_{12}$: $$h_{12} = \frac{1}{a_{12}} = \frac{1}{\frac{13}{12}} = \frac{12}{13}$$ 7. Therefore, the 12th term of the harmonic sequence is $\boxed{\frac{12}{13}}$.