Question 1205173
**1. Set up the Augmented Matrix**

The given system of linear equations can be represented by the following augmented matrix:

```
[ 1   3   2  -1 | -1 ]
[-3  -7  p-6   2 |  1 ]
[ 2   0  p^2   p | q^2 ]
```

**2. Perform Row Operations**

* **R2 = R2 + 3R1:** 
   ```
   [ 1   3   2  -1 | -1 ]
   [ 0   2  p     -1 | -2 ]
   [ 2   0  p^2   p | q^2 ]
   ```

* **R3 = R3 - 2R1:**
   ```
   [ 1   3   2  -1 | -1 ]
   [ 0   2   p     -1 | -2 ]
   [ 0  -6  p^2-4  p+2 | q^2+2 ]
   ```

* **R3 = R3 + 3R2:**
   ```
   [ 1   3   2  -1 | -1 ]
   [ 0   2   p     -1 | -2 ]
   [ 0   0  p^2+3p-4  p-1 | q^2-4 ]
   ```

**3. Analyze for Infinitely Many Solutions**

For the system to have infinitely many solutions with two free variables, the last row of the augmented matrix must be a row of zeros. This implies:

* p^2 + 3p - 4 = 0 
* p - 1 = 0 
* q^2 - 4 = 0

**4. Solve for p and q**

* **p^2 + 3p - 4 = 0** 
   * (p + 4)(p - 1) = 0 
   * p = -4 or p = 1

* **p - 1 = 0** 
   * p = 1

* **q^2 - 4 = 0** 
   * q = ±2

Since we need all conditions to be satisfied simultaneously, the only possible values for p and q are:

* **p = 1** 
* **q = 2 or q = -2**

**5. Solve the System for p = 1, q = 2 (or q = -2)**

* Substitute p = 1 into the row-echelon form of the augmented matrix:
   ```
   [ 1   3   2  -1 | -1 ]
   [ 0   2   1     -1 | -2 ]
   [ 0   0   0     0 | 0 ]
   ```

* Let z = s and w = t (where s and t are free variables).
* From the second row: 2y + s - t = -2 
   => y = (-s + t - 2)/2

* From the first row: x + 3y + 2z - w = -1
   => x + 3[(-s + t - 2)/2] + 2s - t = -1 
   => x = (s - t - 1)/2

**Therefore, the general solution for p = 1 and q = ±2 is:**

* x = (s - t - 1)/2
* y = (-s + t - 2)/2
* z = s
* w = t

where s and t are arbitrary real numbers.

**In summary:**

* The system of linear equations has infinitely many solutions with two free variables only when p = 1 and q = 2 or q = -2. 
* The general solution for these values of p and q is given above.