1. The problem is to express the vector \(\overline{x}\) as a linear combination of vectors \(\overline{p}, \overline{q}, \overline{r}\), i.e., find scalars \(a, b, c\) such that $$\overline{x} = a\overline{p} + b\overline{q} + c\overline{r}.$$ 2. This translates to a system of linear equations: $$a p_1 + b q_1 + c r_1 = x_1,$$ $$a p_2 + b q_2 + c r_2 = x_2,$$ $$a p_3 + b q_3 + c r_3 = x_3,$$ where the subscripts represent the components of the vectors. 3. We form the coefficient matrix and solve for \((a, b, c)\): \[ \begin{pmatrix}p_1 & q_1 & r_1 \\ p_2 & q_2 & r_2 \\ p_3 & q_3 & r_3 \end{pmatrix} \begin{pmatrix}a\\b\\c\end{pmatrix} = \begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} \] 4. Solve using matrix inversion if possible or substitution/row reduction. 5. Specifically for problem 8: \(\overline{x} = (3, 3, -1), \overline{p} = (3, 1, 0), \overline{q} = (-1, 2, 1), \overline{r} = (-1, 0, 2)\) Set up system: $$3a -1b -1c = 3$$ $$1a + 2b + 0c = 3$$ $$0a + 1b + 2c = -1$$ 6. From equation 2: $$a + 2b = 3 \Rightarrow a = 3 - 2b.$$ 7. From equation 3: $$b + 2c = -1 \Rightarrow b = -1 - 2c.$$ 8. Substitute \(a, b\) into equation 1: $$3(3 - 2b) - b - c = 3.$$ Substitute \(b = -1 - 2c\): $$3(3 - 2(-1 - 2c)) - (-1 - 2c) - c = 3.$$ Simplify inside: $$3(3 + 2 + 4c) + 1 + 2c - c = 3,$$ $$3(5 + 4c) + 1 + c = 3,$$ $$15 + 12c + 1 + c = 3,$$ $$16 + 13c = 3,$$ $$13c = 3 - 16 = -13,$$ $$c = -1.$$ 9. Back substitute \(c = -1\) into \(b = -1 - 2c = -1 - 2(-1) = -1 + 2 = 1\). 10. Back substitute \(b = 1\) into \(a = 3 - 2b = 3 - 2(1) = 1\). Answer: $$\boxed{\overline{x} = 1 \cdot \overline{p} + 1 \cdot \overline{q} + (-1) \cdot \overline{r}}.$$