1. **State the problem:** We have the system of equations: $$\begin{cases} x + y + z = 5 \\ x + 2y + z = 9 \\ x + y + (a^2 - 5)z = a \end{cases}$$ We want to find the value(s) of $a$ for which this system has infinitely many solutions. 2. **Analyze the system:** For infinitely many solutions, the system must be consistent and dependent, meaning the third equation must be a linear combination of the first two. 3. **Subtract the first equation from the second:** $$ (x + 2y + z) - (x + y + z) = 9 - 5 \implies y = 4 $$ 4. **Substitute $y=4$ into the first equation:** $$ x + 4 + z = 5 \implies x + z = 1 \implies x = 1 - z $$ 5. **Substitute $x=1 - z$ and $y=4$ into the third equation:** $$ (1 - z) + 4 + (a^2 - 5)z = a $$ Simplify: $$ 5 - z + (a^2 - 5)z = a $$ $$ 5 + z(a^2 - 6) = a $$ 6. **For infinitely many solutions, the third equation must be dependent on the first two, so the coefficient of $z$ must be zero and the constant terms must match:** $$ a^2 - 6 = 0 $$ $$ 5 = a $$ 7. **Solve for $a$:** $$ a^2 = 6 \implies a = \pm \sqrt{6} $$ 8. **Check constants for these $a$ values:** For $a = \sqrt{6}$, the constant term is $a = \sqrt{6}$, but the left side constant is 5, so $5 = \sqrt{6}$ is false. For $a = -\sqrt{6}$, similarly $5 = -\sqrt{6}$ is false. 9. **Conclusion:** The only way for the system to have infinitely many solutions is if the third equation is a linear combination of the first two, which requires both the coefficient of $z$ and the constant term to match. Since the constants do not match for $a = \pm \sqrt{6}$, no infinite solutions occur there. 10. **Check if the system can have infinite solutions for $a=5$:** If $a=5$, then the third equation is: $$ x + y + (25 - 5)z = 5 \implies x + y + 20z = 5 $$ Substitute $x=1 - z$, $y=4$: $$ (1 - z) + 4 + 20z = 5 \implies 5 + 19z = 5 \implies 19z = 0 \implies z=0 $$ Then $x=1$, $y=4$, $z=0$ is a unique solution, not infinite. 11. **Final step:** The system has infinitely many solutions if the third equation is a linear combination of the first two, which happens when: $$ a^2 - 6 = 0 \quad \text{and} \quad a = 5 $$ This is impossible, so no value of $a$ yields infinitely many solutions. **Answer:** There is no value of $a$ for which the system has infinitely many solutions.