= Solution
The <eigenvector> $(1,-1,0)^{\mathsf T}$ has <eigenvalue> $2$. On its <orthogonal complement>, write vectors as $(a,a,c)^{\mathsf T}$. The <eigenvalue equation> becomes
$$
\begin{pmatrix}4&2\\4&4\end{pmatrix}\binom ac=\lambda\binom ac,
\qquad (4-\lambda)^2-8=0.
$$
Hence the ordered <eigenvalues> of the <covariance matrix> are $4+2\sqrt2$, $2$, and $4-2\sqrt2$. The first two <principal components> explain
$$
\boxed{\frac{(4+2\sqrt2)+2}{10}=\frac{3+\sqrt2}{5}\simeq88.28\%}
$$
of the total <variance>. The denominator is the sum of all <eigenvalues>, which equals the <trace>.
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