Answer: C
Let \( \frac{a}{3} = \frac{b}{5} = \frac{c}{8} = k \) Then, \( a = 3k \), \( b = 5k \), \( c = 8k \) Therefore, \[ \frac{3a - 5b + 2c}{a} = \frac{3 \times 3k - 5 \times 5k + 2 \times 8k}{3k} = \frac{9k - 25k + 16k}{3k} = \frac{0}{3k} = 0 \]
Let \( \frac{a}{3} = \frac{b}{5} = \frac{c}{8} = k \) Then, \( a = 3k \), \( b = 5k \), \( c = 8k \) Therefore, \[ \frac{3a - 5b + 2c}{a} = \frac{3 \times 3k - 5 \times 5k + 2 \times 8k}{3k} = \frac{9k - 25k + 16k}{3k} = \frac{0}{3k} = 0 \]