In how many consecutive zeros does the decimal expansion of $\frac{26!}{35^3}$ end?

$\text{(A) }1\qquad\text{(B) }2\qquad\text{(C) }3\qquad\text{(D) }4\qquad\text{(E) }5$
1. **Factorize the given expressions:**
   - First, we need to factorize \(26!\) and \(35^3\).
   - The prime factorization of \(35\) is \(35 = 5 \times 7\). Therefore, \(35^3 = (5 \times 7)^3 = 5^3 \times 7^3\).

2. **Count the factors of 2 and 5 in \(26!\):**
   - To find the number of factors of 2 in \(26!\), we use the formula for the highest power of a prime \(p\) dividing \(n!\):
     \[
     \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \cdots
     \]
     For \(p = 2\) and \(n = 26\):
     \[
     \left\lfloor \frac{26}{2} \right\rfloor + \left\lfloor \frac{26}{4} \right\rfloor + \left\lfloor \frac{26}{8} \right\rfloor + \left\lfloor \frac{26}{16} \right\rfloor = 13 + 6 + 3 + 1 = 23
     \]
     So, there are 23 factors of 2 in \(26!\).

   - To find the number of factors of 5 in \(26!\):
     \[
     \left\lfloor \frac{26}{5} \right\rfloor + \left\lfloor \frac{26}{25} \right\rfloor = 5 + 1 = 6
     \]
     So, there are 6 factors of 5 in \(26!\).

3. **Simplify the fraction \(\frac{26!}{35^3}\):**
   - We have:
     \[
     26! = 2^{23} \times 5^6 \times \text{other primes}
     \]
     \[
     35^3 = 5^3 \times 7^3
     \]
   - Therefore:
     \[
     \frac{26!}{35^3} = \frac{2^{23} \times 5^6 \times \text{other primes}}{5^3 \times 7^3} = 2^{23} \times 5^{6-3} \times \text{other primes} = 2^{23} \times 5^3 \times \text{other primes}
     \]

4. **Determine the number of trailing zeros:**
   - The number of trailing zeros in a number is determined by the number of pairs of factors of 2 and 5.
   - In \(\frac{26!}{35^3}\), we have \(2^{23}\) and \(5^3\). The limiting factor is the number of 5s, which is 3.
   - Therefore, the number of trailing zeros is 3.

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