Apply the orthogonal transformation
$$
D_t=\frac{A_t^+-A_t^-}{\sqrt2},
\qquad
C_t=\frac{A_t^++A_t^-}{\sqrt2}.
$$
The processes $D$ and $C$ are independent one-dimensional Brownian motions, with $D_0=\sqrt2a$ and $C_0=0$. The meeting time is the first time $D$ hits zero, which is almost surely finite by one-dimensional Brownian recurrence.

The <Brownian reflection principle> gives the first-passage density from $x>0$ to zero as
$$
\frac{x}{\sqrt{2\pi u^3}}e^{-x^2/(2u)}.
$$
Substituting $x=\sqrt2a$ gives the <meeting time of two independent Brownian motions> density
$$
\boxed{f_T(u)=\frac a{\sqrt{\pi u^3}}e^{-a^2/u}},
\qquad u>0.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 201 / 4 / c / Solution

Two independent one-dimensional Brownian motions started at $-a$ and $a$ meet at the first time a Brownian motion started at $\sqrt2a$ hits zero. The meeting time has density
$$
f_T(u)=\frac{a}{\sqrt{\pi u^3}}e^{-a^2/u},
\qquad u>0.
$$
An orthogonal sum coordinate is independent of this hitting time and makes the meeting position conditionally $N(0,u/2)$.


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 Incoming links: Meeting time of two independent Brownian motions