1. **State the problem:** Find the values of $c$ such that the angle between vectors $\mathbf{u} = (1,2,1)$ and $\mathbf{v} = (1,0,c)$ is $60^\circ$. 2. **Formula used:** The cosine of the angle $\theta$ between two vectors $\mathbf{u}$ and $\mathbf{v}$ is given by $$\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|}$$ where $\mathbf{u} \cdot \mathbf{v}$ is the dot product and $\|\mathbf{u}\|$ is the magnitude of $\mathbf{u}$. 3. **Calculate dot product:** $$\mathbf{u} \cdot \mathbf{v} = 1 \times 1 + 2 \times 0 + 1 \times c = 1 + 0 + c = 1 + c$$ 4. **Calculate magnitudes:** $$\|\mathbf{u}\| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$$ $$\|\mathbf{v}\| = \sqrt{1^2 + 0^2 + c^2} = \sqrt{1 + c^2}$$ 5. **Set up the equation using $\theta = 60^\circ$:** $$\cos 60^\circ = \frac{1 + c}{\sqrt{6} \sqrt{1 + c^2}}$$ Since $\cos 60^\circ = \frac{1}{2}$, we have $$\frac{1}{2} = \frac{1 + c}{\sqrt{6} \sqrt{1 + c^2}}$$ 6. **Solve for $c$:** Multiply both sides by $\sqrt{6} \sqrt{1 + c^2}$: $$\frac{1}{2} \times \sqrt{6} \sqrt{1 + c^2} = 1 + c$$ 7. **Square both sides to eliminate the square root:** $$\left(\frac{\sqrt{6}}{2} \sqrt{1 + c^2}\right)^2 = (1 + c)^2$$ $$\frac{6}{4} (1 + c^2) = (1 + c)^2$$ $$\frac{3}{2} (1 + c^2) = (1 + c)^2$$ 8. **Expand and simplify:** $$\frac{3}{2} + \frac{3}{2} c^2 = 1 + 2c + c^2$$ 9. **Bring all terms to one side:** $$\frac{3}{2} + \frac{3}{2} c^2 - 1 - 2c - c^2 = 0$$ $$\left(\frac{3}{2} - 1\right) + \left(\frac{3}{2} c^2 - c^2\right) - 2c = 0$$ $$\frac{1}{2} + \frac{1}{2} c^2 - 2c = 0$$ 10. **Multiply entire equation by 2 to clear fractions:** $$1 + c^2 - 4c = 0$$ 11. **Rewrite as a quadratic:** $$c^2 - 4c + 1 = 0$$ 12. **Use quadratic formula:** $$c = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 1 \times 1}}{2} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2}$$ 13. **Simplify $\sqrt{12}$:** $$\sqrt{12} = 2\sqrt{3}$$ 14. **Final solutions:** $$c = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$$ **Answer:** The values of $c$ are $2 + \sqrt{3}$ and $2 - \sqrt{3}$.