= Solution
For the numerical <vectors>, $\mathbf b\times\mathbf c=(-2,1,-2)$ and $D=(-1,1,0)\cdot(-2,1,-2)=3\ne0$, proving noncoplanarity. The <reciprocal basis> is
$$
\boxed{\mathbf A=\frac{(-2,1,-2)}3,\qquad\mathbf B=\frac{(1,1,1)}3,\qquad\mathbf C=\frac{(1,1,-2)}3.}
$$
Reciprocal <dot products> extract the coefficients of any <vector> in the original <basis>. Here $\mathbf A\cdot\mathbf d=-1/3$, $\mathbf B\cdot\mathbf d=2/3$, and $\mathbf C\cdot\mathbf d=5/3$, so
$$
\boxed{\mathbf d=-\frac13\mathbf a+\frac23\mathbf b+\frac53\mathbf c.}
$$
Substitution in Cartesian components recovers $(2,1,-1)$.
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