= Solution
{c}
Put $p=(3x^2-2)/\sqrt2$ and use the <orthonormal basis> $\mathcal B=(1,\sqrt{3/2}\,x,p)$ from <4/b/iii>. The columns of the given <matrix> give
$$
T(1)=-\sqrt2+\frac3{\sqrt2}x^2=p,\qquad T(x)=x,\qquad T(x^2)=-\sqrt2+\sqrt2x^2.
$$
Consequently
$$
T(p)=\frac{3T(x^2)-2T(1)}{\sqrt2}=-1.
$$
The images of $\mathcal B$ are therefore $(p,\sqrt{3/2}\,x,-1)$, again an <orthonormal basis>. By <4/c/i>, \b[$T$ is an <orthogonal transformation>]. Its <matrix> in this <orthonormal basis> is
$$
\boxed{[T]_{\mathcal B}=\begin{pmatrix}0&0&-1\\0&1&0\\1&0&0\end{pmatrix}.}
$$
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