1. **Stating the problem:** We need to find the solution to the system of linear equations (SPL): $$\begin{cases} 2p - 2q - r + 3s = 4 \\ p - q + 2s = 1 \\ -2p + 2q - 4s = -2 \end{cases}$$ 2. **Understanding the system:** We have 3 equations with 4 variables: $p, q, r, s$. Since there are more variables than equations, we expect infinitely many solutions depending on one parameter. 3. **Express variables step-by-step:** From the second equation: $$p - q + 2s = 1 \implies p = q - 2s + 1$$ From the third equation: $$-2p + 2q - 4s = -2$$ Substitute $p$: $$-2(q - 2s + 1) + 2q - 4s = -2$$ Simplify: $$-2q + 4s - 2 + 2q - 4s = -2$$ $$(-2q + 2q) + (4s - 4s) - 2 = -2$$ $$-2 = -2$$ This is always true, so the third equation is dependent on the others. 4. **Use the first equation to find $r$:** $$2p - 2q - r + 3s = 4$$ Substitute $p$: $$2(q - 2s + 1) - 2q - r + 3s = 4$$ Simplify: $$2q - 4s + 2 - 2q - r + 3s = 4$$ $$(2q - 2q) + (-4s + 3s) + 2 - r = 4$$ $$-s + 2 - r = 4$$ Solve for $r$: $$r = 2 - s - 4 = -s - 2$$ 5. **Summary of solutions:** $$p = q - 2s + 1$$ $$r = -s - 2$$ Variables $q$ and $s$ are free parameters. 6. **Final solution:** $$\boxed{\begin{cases} p = q - 2s + 1 \\ r = -s - 2 \\ q = q \text{ (free)} \\ s = s \text{ (free)} \end{cases}}$$ This means the solution set depends on parameters $q$ and $s$.