= Solution
The remaining <eigenvalues> are
$$
\boxed{\lambda_2=\frac35,\qquad\lambda_3=\frac12.}
$$
They have absolute value less than one. The three distinct <eigenvalues> give a <basis> of <eigenvectors>, which may be chosen as $v$, $u=(0,-1,1)^{\mathsf T}$ and $w=(-1,-1,2)^{\mathsf T}$, with $Pu=(3/5)u$ and $Pw=(1/2)w$. Write an arbitrary initial <probability vector> as $x_0=cv+\alpha u+\beta w$. Summing its entries shows $c=1$, because the entries of $v$ sum to one and those of $u,w$ sum to zero. Therefore
$$
P^nx_0=v+\alpha\left(\frac35\right)^nu+\beta\left(\frac12\right)^nw\longrightarrow v.
$$
This proves <convergence> for every initial <probability vector>, rather than merely identifying a possible limit. In particular,
$$
\boxed{\lim_{n\to\infty}x_n=(20\%,45\%,35\%)^{\mathsf T}.}
$$
For the given initial <probability vector>, $\alpha=1/4$ and $\beta=-1/5$; substituting $n=1,2$ also recovers the previous answers. The slowest decaying <eigenvector> component has factor $(3/5)^n$.
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