= Solution
The <orthogonal change of coordinates> preserves distance to the origin. On the required <quadric surface>,
$$
1=-6u^2+3v^2+4w^2\leq4(u^2+v^2+w^2).
$$
Thus the squared distance is at least $1/4$. Equality holds precisely when $u=v=0$, and the constraint then gives $w=\pm1/2$. Using $e_w=(1,0,-3)/\sqrt{10}$ from part (b), the nearest points in the original coordinates are
$$
\boxed{(x,y,z)=\pm\frac{(1,0,-3)}{2\sqrt{10}},\qquad\text{minimum distance }=\frac12.}
$$
Substitution checks the constraint directly: $5x^2-3z^2+6xz=(5-27-18)/40=-1$. The inequality proves these are global nearest points, rather than merely stationary candidates.
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