Let's assume that $T$ is symmetric and see what would happens. We know that there is an orthongonal basis $\{v_1, v_2, v_3\}$ so that
$$Tv_i = \lambda_i v_i.$$
Given that $T(4, 5, 6) = (4, 5, 6)$, we see that $(4, 5, 6)$ is an eigenvector of $T$. Thus we can let
$$v_1 = \frac{(4, 5, 6)}{||(4, 5, 6)||},\ \ \lambda _1= 1.$$
Let $V = \text{span}\langle v_2, v_3\rangle$. Then for all $w = a_2 v_2 + a_3 v_3 \in V$,
$$Tw \cdot v_1 = (a_2Tv_2 + a_3 Tv_3) \cdot v_1 = 0.$$
(As $v_2\cdot v_1 = v_3 \cdot v_1 = 0$). This mean that $Tw \cdot (4, 5, 6) = 0$ whenever $w\in V$. Now consider
$$w = (1, 2, 3) - \big((1, 2, 3)\cdot v_1\big) v_1= (1, 2, 3) - \frac{32}{77} (4, 5, 6)$$
Then $w \in V$ and
$$Tw = T(1, 2, 3) - \frac{32}{77} T(4, 5, 6) = (3, 2, 1) - \frac{32}{77}(4, 5, 6)$$
Thus $Tw \cdot (4, 5, 6) = -4 \neq 0$. This is a contradiction and so $T$ is not symmetric.