We need to find the number of ordered triplets $(a, b, c)$ of positive integers such that $30a + 50b + 70c \leq 343$.

First, we determine the possible values for $c$:

1. Since $70c \leq 343$, we have:
$$c \leq \left\lfloor \frac{343}{70} \right\rfloor = 4$$
Therefore, $c$ can be 1, 2, 3, or 4.

We will consider each case for $c$ and find the corresponding values for $a$ and $b$.

### Case 1: $c = 4$
$$30a + 50b \leq 343 - 70 \cdot 4 = 343 - 280 = 63$$
- Since $30a \geq 30$ and $50b \geq 50$ for positive integers $a$ and $b$, the smallest possible sum $30a + 50b = 30 + 50 = 80$, which is greater than 63. Therefore, there are no solutions in this case.

### Case 2: $c = 3$
$$30a + 50b \leq 343 - 70 \cdot 3 = 343 - 210 = 133$$
- We need to find pairs $(a, b)$ such that $30a + 50b \leq 133$.

- For $b = 1$:
$$30a + 50 \leq 133 \implies 30a \leq 83 \implies a \leq \left\lfloor \frac{83}{30} \right\rfloor = 2$$
Possible pairs: $(1, 1), (2, 1)$

- For $b = 2$:
$$30a + 100 \leq 133 \implies 30a \leq 33 \implies a \leq \left\lfloor \frac{33}{30} \right\rfloor = 1$$
Possible pair: $(1, 2)$

- For $b \geq 3$, $30a + 50b \geq 150$, which is greater than 133. Therefore, no solutions for $b \geq 3$.

Total solutions for $c = 3$: 3 pairs $(1, 1), (2, 1), (1, 2)$.

### Case 3: $c = 2$
$$30a + 50b \leq 343 - 70 \cdot 2 = 343 - 140 = 203$$
- We need to find pairs $(a, b)$ such that $30a + 50b \leq 203$.

- For $b = 1$:
$$30a + 50 \leq 203 \implies 30a \leq 153 \implies a \leq \left\lfloor \frac{153}{30} \right\rfloor = 5$$
Possible pairs: $(1, 1), (2, 1), (3, 1), (4, 1), (5, 1)$

- For $b = 2$:
$$30a + 100 \leq 203 \implies 30a \leq 103 \implies a \leq \left\lfloor \frac{103}{30} \right\rfloor = 3$$
Possible pairs: $(1, 2), (2, 2), (3, 2)$

- For $b = 3$:
$$30a + 150 \leq 203 \implies 30a \leq 53 \implies a \leq \left\lfloor \frac{53}{30} \right\rfloor = 1$$
Possible pair: $(1, 3)$

- For $b \geq 4$, $30a + 50b \geq 200$, which is greater than 203. Therefore, no solutions for $b \geq 4$.

Total solutions for $c = 2$: 9 pairs $(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (1, 2), (2, 2), (3, 2), (1, 3)$.

### Case 4: $c = 1$
$$30a + 50b \leq 343 - 70 \cdot 1 = 343 - 70 = 273$$
- We need to find pairs $(a, b)$ such that $30a + 50b \leq 273$.

- For $b = 1$:
$$30a + 50 \leq 273 \implies 30a \leq 223 \implies a \leq \left\lfloor \frac{223}{30} \right\rfloor = 7$$
Possible pairs: $(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1), (7, 1)$

- For $b = 2$:
$$30a + 100 \leq 273 \implies 30a \leq 173 \implies a \leq \left\lfloor \frac{173}{30} \right\rfloor = 5$$
Possible pairs: $(1, 2), (2, 2), (3, 2), (4, 2), (5, 2)$

- For $b = 3$:
$$30a + 150 \leq 273 \implies 30a \leq 123 \implies a \leq \left\lfloor \frac{123}{30} \right\rfloor = 4$$
Possible pairs: $(1, 3), (2, 3), (3, 3), (4, 3)$

- For $b = 4$:
$$30a + 200 \leq 273 \implies 30a \leq 73 \implies a \leq \left\lfloor \frac{73}{30} \right\rfloor = 2$$
Possible pairs: $(1, 4), (2, 4)$

- For $b = 5$:
$$30a + 250 \leq 273 \implies 30a \leq 23 \implies a \leq \left\lfloor \frac{23}{30} \right\rfloor = 0$$
No solutions for $b = 5$.

Total solutions for $c = 1$: 18 pairs $(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1), (7, 1), (1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (1, 3), (2, 3), (3, 3), (4, 3), (1, 4), (2, 4)$.

Summing up all the solutions from each case:
$$0 + 3 + 9 + 18 = 30$$

The final answer is $\boxed{30}$