1. **State the problem:** We want to find coefficients $a$, $b$, and $c$ such that the system $$ \begin{cases} ax + by - 3z = -24 \\ -2x - by + cz = 42 \\ ax + 3y - cz = -32 \end{cases} $$ has the solution $x=5$, $y=-2$, and $z=6$. 2. **Substitute the solution into each equation:** - For the first equation: $$a(5) + b(-2) - 3(6) = -24$$ Simplify: $$5a - 2b - 18 = -24$$ Add 18 to both sides: $$5a - 2b = -24 + 18$$ $$5a - 2b = -6$$ - For the second equation: $$-2(5) - b(-2) + c(6) = 42$$ Simplify: $$-10 + 2b + 6c = 42$$ Add 10 to both sides: $$2b + 6c = 42 + 10$$ $$2b + 6c = 52$$ - For the third equation: $$a(5) + 3(-2) - c(6) = -32$$ Simplify: $$5a - 6 - 6c = -32$$ Add 6 to both sides: $$5a - 6c = -32 + 6$$ $$5a - 6c = -26$$ 3. **Write the system of equations for $a$, $b$, and $c$:** $$\begin{cases} 5a - 2b = -6 \\ 2b + 6c = 52 \\ 5a - 6c = -26 \end{cases}$$ 4. **Solve the system:** From the first equation: $$5a = -6 + 2b$$ $$a = \frac{-6 + 2b}{5}$$ From the third equation: $$5a - 6c = -26$$ Substitute $a$: $$5 \times \frac{-6 + 2b}{5} - 6c = -26$$ Simplify: $$-6 + 2b - 6c = -26$$ Add 6 to both sides: $$2b - 6c = -20$$ Now we have two equations involving $b$ and $c$: $$\begin{cases} 2b + 6c = 52 \\ 2b - 6c = -20 \end{cases}$$ Add the two equations: $$ (2b + 6c) + (2b - 6c) = 52 + (-20)$$ $$4b = 32$$ $$b = 8$$ Substitute $b=8$ into $2b + 6c = 52$: $$2(8) + 6c = 52$$ $$16 + 6c = 52$$ $$6c = 36$$ $$c = 6$$ Substitute $b=8$ into $a = \frac{-6 + 2b}{5}$: $$a = \frac{-6 + 2(8)}{5} = \frac{-6 + 16}{5} = \frac{10}{5} = 2$$ 5. **Final answer:** $$a = 2, \quad b = 8, \quad c = 6$$