1. The problem is to find the sum of the fractions $\frac{1}{3}$, $\frac{1}{4}$, and $\frac{1}{6}$. 2. To add fractions, we need a common denominator. The denominators are 3, 4, and 6. 3. Find the least common denominator (LCD) of 3, 4, and 6. The prime factors are: - 3 = 3 - 4 = 2^2 - 6 = 2 \times 3 The LCD is $2^2 \times 3 = 12$. 4. Convert each fraction to have denominator 12: $$\frac{1}{3} = \frac{1 \times 4}{3 \times 4} = \frac{4}{12}$$ $$\frac{1}{4} = \frac{1 \times 3}{4 \times 3} = \frac{3}{12}$$ $$\frac{1}{6} = \frac{1 \times 2}{6 \times 2} = \frac{2}{12}$$ 5. Add the fractions: $$\frac{4}{12} + \frac{3}{12} + \frac{2}{12} = \frac{4+3+2}{12} = \frac{9}{12}$$ 6. Simplify the fraction $\frac{9}{12}$ by dividing numerator and denominator by their greatest common divisor 3: $$\frac{\cancel{9}^3}{\cancel{12}^3} = \frac{3}{4}$$ 7. The final answer is $\frac{3}{4}$.