= Solution
By the <spectral theorem for real symmetric matrices>, choose an <orthonormal eigenbasis> $u_1,u_2,u_3$ with $Au_i=\lambda_i u_i$. If $x=\sum_i y_i u_i$, then
$$
1=x^TAx=\sum_i\lambda_i y_i^2
\leq\lambda_1\sum_i y_i^2=\lambda_1|x|^2.
$$
Thus every point of the <ellipsoid> has distance at least $\lambda_1^{-1/2}$ from the origin. Equality holds precisely when
$$
(\lambda_1-\lambda_2)y_2^2+(\lambda_1-\lambda_3)y_3^2=0.
$$
Both coefficients are strictly positive, so $y_2=y_3=0$. The surface equation then gives $y_1=\pm\lambda_1^{-1/2}$. Therefore
$$
\boxed{d_{\min}=\frac1{\sqrt{\lambda_1}},\qquad
x=\pm\frac{u_1}{\sqrt{\lambda_1}}.}
$$
\b[Exactly two points attain the minimum.] The strict ordering makes the largest-eigenvalue <eigenspace> one-dimensional; this is the equality case in <nearest points to the origin on a positive-definite ellipsoid>.
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