= Solution
{c}
Let $B=A^TA$. It is a <symmetric matrix>, and $\mathbf v^TB\mathbf v=\|A\mathbf v\|^2\geq0$. By the <spectral theorem for real symmetric matrices>, choose an <orthonormal basis> $\mathbf v_1,\mathbf v_2,\mathbf v_3$ of <eigenvectors> of $B$, with nonnegative <eigenvalues> $\lambda_1,\lambda_2,\lambda_3$.
For each $i\ne j$, take the unit <parallelogram> parametrized by $s\mathbf v_i+t\mathbf v_j$, $0\leq s,t\leq1$. Its <surface area> is $1$. Moreover,
$$
A\mathbf v_i\cdot A\mathbf v_j=\mathbf v_i^TB\mathbf v_j=0,\qquad
\|A\mathbf v_i\|^2=\lambda_i.
$$
The image <parallelogram> consequently has <area> $\sqrt{\lambda_i\lambda_j}$. By the assumption this equals $1$ for all three pairs, giving
$$
\lambda_1\lambda_2=\lambda_1\lambda_3=\lambda_2\lambda_3=1.
$$
No <eigenvalue> can be zero. Dividing the <equations> and using positivity yields $\lambda_1=\lambda_2=\lambda_3=1$. Hence
$$
\boxed{A^TA=I,\quad\text{so }A\text{ is orthogonal}.}
$$
Thus testing all surfaces is stronger than necessary: the three unit <parallelograms> in these principal directions already determine the result. This is the three-dimensional <characterization of linear surface-area isometries>.
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