To solve the expression \( \frac{15}{34} + \frac{1}{3} + \frac{19}{34} - \frac{3}{4} + \frac{3}{7} \), we need to find a common denominator for all the fractions. The denominators are 34, 3, 4, and 7.

First, let's find the least common multiple (LCM) of these denominators.

- The prime factorization of 34 is \( 2 \times 17 \).
- The prime factorization of 3 is \( 3 \).
- The prime factorization of 4 is \( 2^2 \).
- The prime factorization of 7 is \( 7 \).

The LCM will be the highest power of each prime number that appears in these factorizations:
- The highest power of 2 is \( 2^2 \).
- The highest power of 3 is \( 3 \).
- The highest power of 7 is \( 7 \).
- The highest power of 17 is \( 17 \).

So, the LCM is \( 2^2 \times 3 \times 7 \times 17 = 4 \times 3 \times 7 \times 17 \).

Calculating step-by-step:
- \( 4 \times 3 = 12 \)
- \( 12 \times 7 = 84 \)
- \( 84 \times 17 = 1428 \)

Thus, the common denominator is 1428.

Now, we convert each fraction to have this common denominator:

1. \( \frac{15}{34} \):
\[
\frac{15}{34} = \frac{15 \times 42}{34 \times 42} = \frac{630}{1428}
\]

2. \( \frac{1}{3} \):
\[
\frac{1}{3} = \frac{1 \times 476}{3 \times 476} = \frac{476}{1428}
\]

3. \( \frac{19}{34} \):
\[
\frac{19}{34} = \frac{19 \times 42}{34 \times 42} = \frac{798}{1428}
\]

4. \( \frac{3}{4} \):
\[
\frac{3}{4} = \frac{3 \times 357}{4 \times 357} = \frac{1071}{1428}
\]

5. \( \frac{3}{7} \):
\[
\frac{3}{7} = \frac{3 \times 204}{7 \times 204} = \frac{612}{1428}
\]

Now, we can add and subtract these fractions:
\[
\frac{630}{1428} + \frac{476}{1428} + \frac{798}{1428} - \frac{1071}{1428} + \frac{612}{1428}
\]

Combine the numerators:
\[
630 + 476 + 798 - 1071 + 612 = 1445
\]

So, the result is:
\[
\frac{1445}{1428}
\]

This fraction can be simplified. The greatest common divisor (GCD) of 1445 and 1428 is 1, so the fraction is already in its simplest form.

Thus, the final answer is:
\[
\frac{1445}{1428}
\]

Or, as a mixed number:
\[
1 \frac{17}{1428}
\]