1. Interpret the condition

We are given

with $a,b,c\neq 0$, and $A^T A=I.$

Since

we have $A^T=A$, so the condition becomes $A^2=I.$

Also given: $a^3+b^3+c^3=2.$

We need to find a possible value of $abc$.

2. Use $A^2=I$

Let us compute entries of $A^2$.

The $(1,1)$ entry is $a^2+b^2+c^2.$ Since $A^2=I$, this must equal $1$: $a^2+b^2+c^2=1. \qquad (1)$

The $(1,2)$ entry is $ab+bc+ca.$ Since off-diagonal entries of $I$ are zero, $ab+bc+ca=0. \qquad (2)$

(Other off-diagonal entries give the same condition by symmetry.)

3. Use the identity for sum of cubes

We know $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca).$

From (1) and (2), $a^2+b^2+c^2-ab-bc-ca=1-0=1.$ So $a^3+b^3+c^3-3abc=a+b+c. \qquad (3)$

But from (2), $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=1+0=1.$ Hence $a+b+c=\pm 1. \qquad (4)$

Given $a^3+b^3+c^3=2$, substitute into (3): $2-3abc=a+b+c.$ Using (4), we get two cases:

• If $a+b+c=1$, then $2-3abc=1 \implies 3abc=1 \implies abc=\frac13.$

• If $a+b+c=-1$, then $2-3abc=-1 \implies 3abc=3 \implies abc=1.$

So algebraically, $abc\in\left\{\frac13,1\right\}.$

4. Check which value is among the options

The options are:

• A: $3$
• B: $\frac13$
• C: $-\frac13$
• D: $\frac23$

Among these, only $\frac13$ is possible.

5. Answer

Therefore, the correct option is $\boxed{\frac13}.$

So Option B is correct.