What is the remainder, in base $10$, when $24_7 + 364_7 + 43_7 + 12_7 + 3_7 + 1_7$ is divided by $6$?

1. Convert each base \(7\) number to base \(10\):
   - \(24_7\):
     \[
     24_7 = 2 \cdot 7^1 + 4 \cdot 7^0 = 2 \cdot 7 + 4 = 14 + 4 = 18_{10}
     \]
   - \(364_7\):
     \[
     364_7 = 3 \cdot 7^2 + 6 \cdot 7^1 + 4 \cdot 7^0 = 3 \cdot 49 + 6 \cdot 7 + 4 = 147 + 42 + 4 = 193_{10}
     \]
   - \(43_7\):
     \[
     43_7 = 4 \cdot 7^1 + 3 \cdot 7^0 = 4 \cdot 7 + 3 = 28 + 3 = 31_{10}
     \]
   - \(12_7\):
     \[
     12_7 = 1 \cdot 7^1 + 2 \cdot 7^0 = 1 \cdot 7 + 2 = 7 + 2 = 9_{10}
     \]
   - \(3_7\):
     \[
     3_7 = 3 \cdot 7^0 = 3_{10}
     \]
   - \(1_7\):
     \[
     1_7 = 1 \cdot 7^0 = 1_{10}
     \]

2. Sum the base \(10\) equivalents:
   \[
   18 + 193 + 31 + 9 + 3 + 1 = 255
   \]

3. Find the remainder when \(255\) is divided by \(6\):
   \[
   255 \div 6 = 42 \text{ remainder } 3
   \]
   Therefore, the remainder is \(3\).

The final answer is \(\boxed{3}\).