= Solution
Place observations in rows of the data <matrix> $X$. Since the coordinate <sample means> are zero, the <Gram matrix> of its columns is
$$
X^TX=\begin{pmatrix}114&-48\\-48&86\end{pmatrix}.
$$
The <sample covariance matrix> is $S=X^TX/(3-1)$; using divisor three instead changes <eigenvalues> by a common factor and leaves <principal component loadings> unchanged. For $X^TX$, the <characteristic polynomial> is
$$
\det(X^TX-\lambda I)=(114-\lambda)(86-\lambda)-48^2
=\lambda^2-200\lambda+7500=(\lambda-150)(\lambda-50).
$$
For the larger <eigenvalue> $150$, the <eigenvector> equation gives $-36\alpha-48\beta=0$, so $3\alpha+4\beta=0$. Normalize to unit <Euclidean norm> and choose $v_1=(4/5,-3/5)^T$. An orthogonal unit <eigenvector> for $50$ is $v_2=(3/5,4/5)^T$. Therefore
$$
\boxed{PC_1=\frac45x-\frac35y,\qquad PC_2=\frac35x+\frac45y}.
$$
The <sample variances> of these <principal components> are $75$ and $25$, respectively. Their <principal component scores> for observations $A,B,C$ are
$$
\begin{array}{c|rr}
&PC_1&PC_2\\\hline
A&5&5\\
B&-10&0\\
C&5&-5
\end{array}.
$$
Either <eigenvector> may be multiplied by $-1$ without changing its <principal component> direction or its <variance>. No coordinate standardization is applied here: that would instead be <principal component analysis on a correlation matrix>.
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