If \[ \begin {eqnarray*} x + y + z + w = 20 \\ y + 2z - 3w = 28 \\ x - 2y + z = 36 \\ -7x - y + 5z + 3w = 84 \]then what is $(x,y,z,w)$?
1. We start with the given system of equations:
   \[
   \begin{cases}
   x + y + z + w = 20 \\
   y + 2z - 3w = 28 \\
   x - 2y + z = 36 \\
   -7x - y + 5z + 3w = 84
   \end{cases}
   \]

2. Add the second and fourth equations:
   \[
   y + 2z - 3w + (-7x - y + 5z + 3w) = 28 + 84
   \]
   Simplifying, we get:
   \[
   -7x + 7z = 112 \implies -x + z = 16 \implies x = z - 16
   \]

3. Substitute \( x = z - 16 \) into the third equation:
   \[
   (z - 16) - 2y + z = 36
   \]
   Simplifying, we get:
   \[
   2z - 2y - 16 = 36 \implies 2z - 2y = 52 \implies z - y = 26 \implies y = z - 26
   \]

4. Substitute \( x = z - 16 \) and \( y = z - 26 \) into the first equation:
   \[
   (z - 16) + (z - 26) + z + w = 20
   \]
   Simplifying, we get:
   \[
   3z - 42 + w = 20 \implies 3z + w = 62
   \]

5. Substitute \( y = z - 26 \) into the second equation:
   \[
   (z - 26) + 2z - 3w = 28
   \]
   Simplifying, we get:
   \[
   3z - 26 - 3w = 28 \implies 3z - 3w = 54 \implies z - w = 18 \implies w = z - 18
   \]

6. Now we have two equations:
   \[
   \begin{cases}
   3z + w = 62 \\
   w = z - 18
   \end{cases}
   \]
   Substitute \( w = z - 18 \) into \( 3z + w = 62 \):
   \[
   3z + (z - 18) = 62
   \]
   Simplifying, we get:
   \[
   4z - 18 = 62 \implies 4z = 80 \implies z = 20
   \]

7. Using \( z = 20 \), we find:
   \[
   x = z - 16 = 20 - 16 = 4
   \]
   \[
   y = z - 26 = 20 - 26 = -6
   \]
   \[
   w = z - 18 = 20 - 18 = 2
   \]

Thus, the solution to the system of equations is:
\[
(x, y, z, w) = (4, -6, 20, 2)
\]

The final answer is \(\boxed{(4, -6, 20, 2)}\)