Let $a,b,c\in \mathbb{R}$ with $a^2+b^2+c^2=1$ and $\lambda\in \mathbb{R}_{>0}\setminus\{1\}$. Then for each solution $(x,y,z)$ of the system of equations:
\[
\begin{cases}
x-\lambda y=a,\\
y-\lambda z=b,\\
z-\lambda x=c.
\end{cases}
\] 
we have $\displaystyle x^2+y^2+z^2\leqslant \frac1{(\lambda-1)^2}$.

[i]Radu Gologan[/i]
1. Given the system of equations:
   \[
   \begin{cases}
   x - \lambda y = a, \\
   y - \lambda z = b, \\
   z - \lambda x = c,
   \end{cases}
   \]
   we need to find the solutions for \(x\), \(y\), and \(z\).

2. Start by expressing \(x\) in terms of \(y\) and \(z\):
   \[
   x = a + \lambda y.
   \]

3. Substitute \(y\) from the second equation:
   \[
   y = b + \lambda z.
   \]

4. Substitute \(z\) from the third equation:
   \[
   z = c + \lambda x.
   \]

5. Substitute \(z\) back into the expression for \(y\):
   \[
   y = b + \lambda (c + \lambda x) = b + \lambda c + \lambda^2 x.
   \]

6. Substitute \(y\) back into the expression for \(x\):
   \[
   x = a + \lambda (b + \lambda c + \lambda^2 x) = a + \lambda b + \lambda^2 c + \lambda^3 x.
   \]

7. Solve for \(x\):
   \[
   x - \lambda^3 x = a + \lambda b + \lambda^2 c \implies x(1 - \lambda^3) = a + \lambda b + \lambda^2 c \implies x = \frac{a + \lambda b + \lambda^2 c}{1 - \lambda^3}.
   \]

8. Similarly, solve for \(y\) and \(z\):
   \[
   y = \frac{b + \lambda c + \lambda^2 a}{1 - \lambda^3},
   \]
   \[
   z = \frac{c + \lambda a + \lambda^2 b}{1 - \lambda^3}.
   \]

9. Now, calculate \(x^2 + y^2 + z^2\):
   \[
   x^2 + y^2 + z^2 = \left( \frac{a + \lambda b + \lambda^2 c}{1 - \lambda^3} \right)^2 + \left( \frac{b + \lambda c + \lambda^2 a}{1 - \lambda^3} \right)^2 + \left( \frac{c + \lambda a + \lambda^2 b}{1 - \lambda^3} \right)^2.
   \]

10. Factor out the common denominator:
    \[
    x^2 + y^2 + z^2 = \frac{(a + \lambda b + \lambda^2 c)^2 + (b + \lambda c + \lambda^2 a)^2 + (c + \lambda a + \lambda^2 b)^2}{(1 - \lambda^3)^2}.
    \]

11. Expand the numerators:
    \[
    (a + \lambda b + \lambda^2 c)^2 + (b + \lambda c + \lambda^2 a)^2 + (c + \lambda a + \lambda^2 b)^2.
    \]

12. Use the identity \(a^2 + b^2 + c^2 = 1\) and the symmetry of the terms:
    \[
    (a + \lambda b + \lambda^2 c)^2 + (b + \lambda c + \lambda^2 a)^2 + (c + \lambda a + \lambda^2 b)^2 = (a^2 + b^2 + c^2)(1 + \lambda^2 + \lambda^4) + 2(ab + ac + bc)(\lambda + \lambda^2 + \lambda^3).
    \]

13. Substitute \(a^2 + b^2 + c^2 = 1\):
    \[
    = 1 \cdot (1 + \lambda^2 + \lambda^4) + 2(ab + ac + bc)(\lambda + \lambda^2 + \lambda^3).
    \]

14. Apply the Cauchy-Schwarz inequality to bound \(ab + ac + bc\):
    \[
    ab + ac + bc \leq \sqrt{(a^2 + b^2 + c^2)(a^2 + b^2 + c^2)} = 1.
    \]

15. Thus:
    \[
    x^2 + y^2 + z^2 \leq \frac{1 \cdot (1 + \lambda^2 + \lambda^4) + 2 \cdot 1 \cdot (\lambda + \lambda^2 + \lambda^3)}{(1 - \lambda^3)^2}.
    \]

16. Simplify the numerator:
    \[
    = \frac{1 + \lambda^2 + \lambda^4 + 2\lambda + 2\lambda^2 + 2\lambda^3}{(1 - \lambda^3)^2} = \frac{1 + 3\lambda^2 + 2\lambda + 2\lambda^3 + \lambda^4}{(1 - \lambda^3)^2}.
    \]

17. Notice that:
    \[
    1 + 3\lambda^2 + 2\lambda + 2\lambda^3 + \lambda^4 = (1 + \lambda + \lambda^2)^2.
    \]

18. Therefore:
    \[
    x^2 + y^2 + z^2 \leq \frac{(1 + \lambda + \lambda^2)^2}{(1 - \lambda^3)^2}.
    \]

19. Since \((1 + \lambda + \lambda^2) = (1 - \lambda + \lambda^2)(1 + \lambda + \lambda^2)\):
    \[
    x^2 + y^2 + z^2 \leq \frac{1}{(1 - \lambda)^2}.
    \]

The final answer is \( \boxed{ \frac{1}{(1 - \lambda)^2} } \).