Are there non-commuting matrices for which $\mathrm{tr}(ABC)=\mathrm{tr}(BAC)$?

Given $C$ and $A$, $$\varphi(B) = \text{tr}(ABC) - \text{tr}(BAC) = \text{tr}(B (CA-AC))$$ is a linear functional on $M_{n\times n}(\mathbb C)$, and if not identically $0$ it is $0$ on a linear subspace of $M_{n \times n}(\mathbb C)$ of codimension $1$. Thus for any $B_1$ and $B_2$ with $\varphi(B_2) \ne 0$, there is some $t$ such that $\varphi(B_1 + t B_2) = 0$ (namely $t = -\varphi(B_1)/\varphi(B_2)$).

it's worth mentioning an important special case that comes up e.g. in semi-definite programming.
Suppose $A,B,C$ are arbitrary real symmetric matrices, then

$\text{trace}\Big(ABC\Big)$
$=\text{trace}\Big(BCA\Big)$
$=\text{trace}\Big(\big(BCA\big)^T\Big)$
$=\text{trace}\Big(A^TC^TB^T\Big)$
$=\text{trace}\Big(ACB\Big)$
$=\text{trace}\Big(BAC\Big)$

(This proof of course formally works the same when all 3 are complex symmetric)