Let $\tau_a$ denote the first passage time to $a>0$ for a Brownian motion $W(t)$. Then is the claimed identity true?
$$ \mathbb{P}(\tau_a\leq \tau_{-a}\leq t)=\mathbb{P}(\tau_{3a}\leq t) $$
I had a heuristic (diagram) idea through reflection principle, but I am not sure if it is correct. The form of reflection principle that I am using is, $$ \mathbb{P}(\tau_a\leq t, W(t)\leq b)=\mathbb{P}(W(t)\geq 2a-b) $$ where $b\leq a$. I am not able to rigorously prove or disprove the identity above since I am not able to deal with the joint distribution of the first passage time with the reflection principle.
The claimed identity is not true. Rather than trying to use an identity derived from the reflection principle, it is better to use the reflection principle directly. Let $\gamma_{-a}:=\min\{t>\tau_a: W_t=-a\}$. Then applying reflection at time $\tau_a$ shows that $$ \mathbb{P}(\tau_a\leq \gamma_{-a}\leq t)=\mathbb{P}(\tau_{3a}\leq t) \,. \tag{1} $$
However, applying reflection at time $\tau_a$ also shows that
$$ \mathbb{P}(\tau_a\leq \tau_{-a}\leq t)=\mathbb{P}(\tau_a\leq \tau_{-a} \quad \text{and} \quad \tau_{3a}\leq t) \,. \tag{2} $$ Incidentally, we can obtain a lower bound for the difference between (1) and (2) by applying the Markov property at time $\tau_a$:
$$\mathbb{P}(\tau_{3a}\leq t)-\mathbb{P}(\tau_a\leq \tau_{-a} \quad \text{and} \quad \tau_{3a}\leq t) =$$ $$ =\mathbb{P}(\tau_a>\tau_{-a} \quad \text{and} \quad \tau_{3a}\leq t) \ge $$ $$ \ge \mathbb{P}(t/2 \ge \tau_a> \tau_{-a} \quad \text{and} \quad \tau_{3a}\leq \tau_a+t/2) =$$ $$ = \mathbb{P}(t/2 \ge \tau_a> \tau_{-a}) \cdot \mathbb{P}(\tau_{2a} \le t/2) \ge $$ $$\ge \mathbb{P}( \tau_{-3a} \le t/2) \cdot \mathbb{P}(\tau_{2a} \le t/2)\,.$$ (using reflection at time $\tau_{-a}$ in the last step.)