Step 1: First count all perfect-square factors of \( N=2^{11}3^{11}5^{9}7^{10} \), ignoring the 420 requirement. For each prime, the count of even exponents from 0 up to its cap is \( \lfloor\text{cap}/2\rfloor+1 \): prime 2 gives 6, prime 3 gives 6, prime 5 gives 5, prime 7 gives 6. Total \( =6\times6\times5\times6=1080 \).
Step 2: Now remove the ones that fail the 420 floor. For prime 2 (needs exponent at least 2), only exponent 0 is even and too small, one bad case out of 6, leaving 5. For prime 3 (needs at least 1), only exponent 0 is even and too small, leaving 5 out of 6. Same reasoning for prime 5, leaving 4 out of 5, and prime 7, leaving 5 out of 6.
Step 3: Multiply the surviving counts:
\[ 5\times5\times4\times5 = \boxed{500}. \]
Final Answer: 500 factors qualify. (Option 3)