1. **State the problem:** Solve the equation $x^2 + y^2 + z^2 = 3xyz$ for real numbers $x, y, z$. 2. **Recognize the form:** This is a symmetric equation in three variables. One common approach is to check for solutions where variables are equal or zero. 3. **Try the case $x = y = z = t$:** $$t^2 + t^2 + t^2 = 3t \, t \, t \implies 3t^2 = 3t^3$$ Simplify: $$3t^2 = 3t^3 \implies \cancel{3}t^2 = \cancel{3}t^3 \implies t^2 = t^3$$ Divide both sides by $t^2$ (assuming $t \neq 0$): $$\frac{t^2}{\cancel{t^2}} = \frac{t^3}{\cancel{t^2}} \implies 1 = t$$ So $t = 1$ is a solution. 4. **Check $t=0$:** If $t=0$, then $0 = 0$ holds, so $(0,0,0)$ is also a solution. 5. **Summary of equal variables case:** Solutions are $(0,0,0)$ and $(1,1,1)$. 6. **Try if any variable is zero, say $z=0$:** Equation becomes: $$x^2 + y^2 + 0 = 3xy \cdot 0 = 0$$ So: $$x^2 + y^2 = 0$$ This implies $x=0$ and $y=0$. So $(0,0,0)$ again. 7. **Try to find other solutions:** Rewrite the equation as: $$x^2 + y^2 + z^2 - 3xyz = 0$$ This is a known form related to the Markov equation. The integer solutions are well-studied. 8. **Check for solutions with one variable equal to 1:** Let $z=1$, then: $$x^2 + y^2 + 1 = 3xy$$ Rearranged: $$x^2 - 3xy + y^2 + 1 = 0$$ Consider this as a quadratic in $x$: $$x^2 - 3y x + (y^2 + 1) = 0$$ Discriminant: $$\Delta = ( -3y)^2 - 4(y^2 + 1) = 9y^2 - 4y^2 - 4 = 5y^2 - 4$$ For real $x$, $\Delta \geq 0$: $$5y^2 - 4 \geq 0 \implies y^2 \geq \frac{4}{5}$$ So for $|y| \geq \frac{2}{\sqrt{5}}$, real $x$ exist. 9. **Final note:** The equation has infinite real solutions, including $(0,0,0)$, $(1,1,1)$, and others satisfying the above conditions. **Final answer:** $$\boxed{(0,0,0), (1,1,1), \text{and infinitely many other real triples } (x,y,z) \text{ satisfying } x^2 + y^2 + z^2 = 3xyz}$$