= Solution
For $\lambda=5$, the <eigenvalue equation> reduces to a one-dimensional <eigenspace> spanned by $(1,2,-1)$. For $\lambda=-3$, the rows of $A+3I$ are multiples of $(1,2,-3)$, so its <eigenspace> is the <plane> $x+2y-3z=0$. One possible set of three independent <eigenvectors> is
$$
\boxed{\mathbf e_1=(1,2,-1)\quad(\lambda=5),\qquad
\mathbf e_2=(-2,1,0),\quad\mathbf e_3=(3,0,1)\quad(\lambda=-3)}.
$$
Every <eigenvector> for $-3$ is a nonzero <linear combination> of the last two. Their two-dimensional <eigenspace> shows that the repeated root is not defective: the <matrix> is <diagonalizable>.
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