Find all $7$-digit numbers which are multiples of $21$ and which have each digit $3$ or $7$.	
To find all 7-digit numbers that are multiples of 21 and have each digit as either 3 or 7, we need to consider the properties of numbers divisible by 21. A number is divisible by 21 if and only if it is divisible by both 3 and 7.

1. **Divisibility by 3:**
   A number is divisible by 3 if the sum of its digits is divisible by 3. Since each digit is either 3 or 7, we need to consider the possible sums of the digits:
   - If all digits are 3, the sum is \(7 \times 3 = 21\), which is divisible by 3.
   - If there are four 3s and three 7s, the sum is \(4 \times 3 + 3 \times 7 = 12 + 21 = 33\), which is divisible by 3.
   - If there is one 3 and six 7s, the sum is \(1 \times 3 + 6 \times 7 = 3 + 42 = 45\), which is divisible by 3.

2. **Divisibility by 7:**
   We need to check which of the above cases are also divisible by 7. 

   - **Case 1: Seven 3s**
     The number is \(3333333\). We need to check if this number is divisible by 7. However, since \(7\) does not divide \(1111111\), this case does not work.

   - **Case 2: Four 3s and three 7s**
     We need to find combinations of four 3s and three 7s such that the resulting number is divisible by 7. We can write the number in the form \(10^a + 10^b + 10^c + 10^d\) modulo 7 for some distinct \(a, b, c, d\) in \(\{0, 1, 2, 3, 4, 5, 6\}\).

     The powers of 10 modulo 7 are:
     \[
     10^0 \equiv 1 \mod 7, \quad 10^1 \equiv 3 \mod 7, \quad 10^2 \equiv 2 \mod 7, \quad 10^3 \equiv 6 \mod 7, \quad 10^4 \equiv 4 \mod 7, \quad 10^5 \equiv 5 \mod 7, \quad 10^6 \equiv 1 \mod 7
     \]

     We need to find sets of four distinct indices such that their sum modulo 7 is 0. The possible sets are:
     \[
     \{2, 3, 5, 6\}, \{0, 2, 3, 5\}, \{0, 1, 3, 4\}, \{1, 3, 4, 6\}, \{1, 2, 4, 5\}, \{0, 1, 2, 6\}
     \]

     These correspond to the following numbers:
     \[
     3373377, 7373373, 7733733, 3733737, 7337337, 3777333
     \]

   - **Case 3: One 3 and six 7s**
     The number can be written as \(7777777 - 4 \times 10^k\) for \(k = 0, 2, \ldots, 6\). If this is a multiple of 7, then so is \(4 \times 10^k\). However, since 4 is not divisible by 7, this case does not work.

The final answer is \( \boxed{ \{3373377, 7373373, 7733733, 3733737, 7337337, 3777333\} } \)