Let $$a, b, c \in R$$ be all non-zero and satisfies $$a^3 + b^3 + c^3 = 2$$. If the matrix $$A = \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix}$$ satisfies $$A^TA = I$$, then a value of $$abc$$ can be:

We begin with the information that the real, non-zero numbers $$a,\;b,\;c$$ are placed in the matrix

$$A=\begin{bmatrix}a&b&c\\[2pt]b&c&a\\[2pt]c&a&b\end{bmatrix}$$

and that this matrix satisfies $$A^{T}A=I_{3}$$. The columns of an orthogonal matrix are orthonormal, so the dot-product of a column with itself equals $$1$$ and the dot-product of two different columns equals $$0$$. Writing out these conditions gives

$$\begin{aligned} &(1)\quad a^{2}+b^{2}+c^{2}=1,\\[4pt] &(2)\quad ab+bc+ca=0. \end{aligned}$$

We shall denote the elementary symmetric sums by

$$S_{1}=a+b+c,\qquad S_{2}=ab+bc+ca,\qquad S_{3}=abc.$$

From (2) we already have $$S_{2}=0$$. Squaring $$S_{1}$$ and using (1) and (2) we get

$$S_{1}^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca)=1+2\cdot0=1,$$

so

$$S_{1}=a+b+c=\pm1.\qquad(3)$$

Next we employ the well-known identity

$$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)\bigl(a^{2}+b^{2}+c^{2}-ab-bc-ca\bigr).$$

Because of (1) and (2) the factor in parentheses is simply $$1$$, therefore

$$a^{3}+b^{3}+c^{3}-3abc=a+b+c=S_{1}.$$

The question tells us that $$a^{3}+b^{3}+c^{3}=2$$, so substituting this value we obtain the linear relation

$$2-3S_{3}=S_{1}.\qquad(4)$$

Using (3) inside (4) we analyse the two possible signs separately.

• If $$S_{1}=1$$ then $$2-3S_{3}=1\Longrightarrow3S_{3}=1\Longrightarrow S_{3}=abc=\dfrac13.$$

• If $$S_{1}=-1$$ then $$2-3S_{3}=-1\Longrightarrow3S_{3}=3\Longrightarrow S_{3}=abc=1.$$

So the algebra allows two numerical values for $$abc$$, namely $$\dfrac13$$ and $$1$$.

However, the four options offered in the problem statement are

$$-\dfrac13,\;\dfrac13,\;3,\;\dfrac23,$$

among which only $$\dfrac13$$ appears. Consequently, that is the admissible value demanded by the question.

Hence, the correct answer is Option B.