= Solution
Find a <stationary distribution> $\boldsymbol\pi=(u,v,w)^T$ from $P\boldsymbol\pi=\boldsymbol\pi$ and $u+v+w=1$. The first two independent <linear equations> are
$$
4u-2v-w=0,\qquad -2u+3v-3w=0.
$$
They give $u=9w/8$ and $v=7w/4$. Normalization gives $w=8/31$, hence
$$
\boxed{\boldsymbol\pi=\frac1{31}\begin{pmatrix}9\\14\\8\end{pmatrix},\qquad
P^n\begin{pmatrix}1\\0\\0\end{pmatrix}\longrightarrow\boldsymbol\pi.}
$$
To justify the limit using the assumptions, distinct <eigenvalues> give an <eigenbasis>, over the <complex numbers> if necessary. Choose $\boldsymbol\pi$ as the <eigenvector> for 1 and let $\mathbf v_2,\mathbf v_3$ correspond to the other <eigenvalues> $\lambda_2,\lambda_3$. Expand the initial <probability distribution> as
$$
\mathbf p_0=\alpha\boldsymbol\pi+\beta\mathbf v_2+\gamma\mathbf v_3.
$$
Since $\mathbf1^TP=\mathbf1^T$, each $\lambda_j\ne1$ satisfies
$$
\mathbf1^T\mathbf v_j=\mathbf1^TP\mathbf v_j
=\lambda_j\mathbf1^T\mathbf v_j,
\qquad\text{so}\qquad\mathbf1^T\mathbf v_j=0.
$$
Multiplying the expansion by $\mathbf1^T$ therefore gives $\alpha=1$. It follows that
$$
P^n\mathbf p_0=\boldsymbol\pi+\beta\lambda_2^n\mathbf v_2
+\gamma\lambda_3^n\mathbf v_3\longrightarrow\boldsymbol\pi,
$$
because $|\lambda_2|,|\lambda_3|<1$. The components of the limiting <probability distribution> are approximately $0.2903,0.4516,0.2581$.
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