Let $ p_1, p_2, p_3$ and $ p_4$ be four different prime numbers satisying the equations

                   $ 2p_1 \plus{} 3p_2 \plus{} 5p_3 \plus{} 7p_4 \equal{} 162$
                 $ 11p_1 \plus{} 7p_2 \plus{} 5p_3 \plus{} 4p_4 \equal{} 162$

Find all possible values of the product $ p_1p_2p_3p_4$
1. **Subtract the two given equations:**

   \[
   2p_1 + 3p_2 + 5p_3 + 7p_4 = 162
   \]
   \[
   11p_1 + 7p_2 + 5p_3 + 4p_4 = 162
   \]

   Subtracting the second equation from the first:

   \[
   (2p_1 + 3p_2 + 5p_3 + 7p_4) - (11p_1 + 7p_2 + 5p_3 + 4p_4) = 0
   \]

   Simplifying, we get:

   \[
   2p_1 + 3p_2 + 7p_4 - 11p_1 - 7p_2 - 4p_4 = 0
   \]

   \[
   -9p_1 - 4p_2 + 3p_4 = 0
   \]

   \[
   9p_1 + 4p_2 = 3p_4
   \]

   \[
   3p_4 = 9p_1 + 4p_2
   \]

   \[
   p_4 = 3p_1 + \frac{4p_2}{3}
   \]

2. **Determine \( p_2 \):**

   Since \( p_4 \) must be a prime number, \( \frac{4p_2}{3} \) must be an integer. This implies that \( p_2 \) must be divisible by 3. The only prime number divisible by 3 is 3 itself. Therefore, \( p_2 = 3 \).

3. **Substitute \( p_2 = 3 \) into the equations:**

   \[
   2p_1 + 3(3) + 5p_3 + 7p_4 = 162
   \]

   \[
   2p_1 + 9 + 5p_3 + 7p_4 = 162
   \]

   \[
   2p_1 + 5p_3 + 7p_4 = 153 \quad \text{(Equation 1)}
   \]

   \[
   11p_1 + 7(3) + 5p_3 + 4p_4 = 162
   \]

   \[
   11p_1 + 21 + 5p_3 + 4p_4 = 162
   \]

   \[
   11p_1 + 5p_3 + 4p_4 = 141 \quad \text{(Equation 2)}
   \]

4. **Subtract the new equations:**

   \[
   (2p_1 + 5p_3 + 7p_4) - (11p_1 + 5p_3 + 4p_4) = 153 - 141
   \]

   \[
   2p_1 + 7p_4 - 11p_1 - 4p_4 = 12
   \]

   \[
   -9p_1 + 3p_4 = 12
   \]

   \[
   -3p_1 + p_4 = 4
   \]

   \[
   p_4 = 3p_1 + 4
   \]

5. **Determine \( p_1 \) and \( p_4 \):**

   Since \( p_4 \) must be a prime number, we need to find \( p_1 \) such that \( 3p_1 + 4 \) is a prime number. Let's test small prime numbers for \( p_1 \):

   - If \( p_1 = 2 \):
     \[
     p_4 = 3(2) + 4 = 6 + 4 = 10 \quad \text{(not a prime)}
     \]

   - If \( p_1 = 3 \):
     \[
     p_4 = 3(3) + 4 = 9 + 4 = 13 \quad \text{(prime)}
     \]

   Therefore, \( p_1 = 3 \) and \( p_4 = 13 \).

6. **Determine \( p_3 \):**

   Substitute \( p_1 = 3 \), \( p_2 = 3 \), and \( p_4 = 13 \) into one of the original equations to find \( p_3 \):

   \[
   2(3) + 3(3) + 5p_3 + 7(13) = 162
   \]

   \[
   6 + 9 + 5p_3 + 91 = 162
   \]

   \[
   106 + 5p_3 = 162
   \]

   \[
   5p_3 = 56
   \]

   \[
   p_3 = \frac{56}{5} \quad \text{(not an integer)}
   \]

   This indicates an error in our previous steps. Let's re-evaluate the possible values for \( p_1 \).

   - If \( p_1 = 5 \):
     \[
     p_4 = 3(5) + 4 = 15 + 4 = 19 \quad \text{(prime)}
     \]

   Therefore, \( p_1 = 5 \) and \( p_4 = 19 \).

7. **Recalculate \( p_3 \):**

   Substitute \( p_1 = 5 \), \( p_2 = 3 \), and \( p_4 = 19 \) into one of the original equations to find \( p_3 \):

   \[
   2(5) + 3(3) + 5p_3 + 7(19) = 162
   \]

   \[
   10 + 9 + 5p_3 + 133 = 162
   \]

   \[
   152 + 5p_3 = 162
   \]

   \[
   5p_3 = 10
   \]

   \[
   p_3 = 2
   \]

8. **Calculate the product \( p_1p_2p_3p_4 \):**

   \[
   p_1 = 5, \quad p_2 = 3, \quad p_3 = 2, \quad p_4 = 19
   \]

   \[
   p_1p_2p_3p_4 = 5 \times 3 \times 2 \times 19 = 570
   \]

The final answer is \( \boxed{570} \).