Let $\boldsymbol{\xi} = (\xi_1,\xi_2,\xi_3)$ such that $\xi_i\geq 0$and $\xi_1+\xi_2+\xi_3 = 1$. Let $Y\sim N_3(\boldsymbol{\mu}(\boldsymbol{\xi}), \mathrm{\Sigma}(\boldsymbol{\xi}))$, where the components of the covariance matrix (which is singular in our case) are function of $\boldsymbol{\xi}$
\begin{equation}
\boldsymbol{\mu}(\boldsymbol{\xi}) = \begin{bmatrix}
(\mu_1-\mu_2)\sqrt{\frac{\xi_1\xi_2}{\xi_1+\xi_2}}\\
(\mu_1-\mu_3)\sqrt{\frac{\xi_1\xi_3}{\xi_1+\xi_3}}\\
(\mu_2-\mu_3)\sqrt{\frac{\xi_2\xi_3}{\xi_2+\xi_3}}\end{bmatrix}.
\end{equation}
and
\begin{equation}
\mathrm{\Sigma}(\boldsymbol{\xi}) = \begin{bmatrix}
1 & \sqrt{\frac{\xi_2\xi_3}{(\xi_2+\xi_1)(\xi_3+\xi_1)}} & -\sqrt{\frac{\xi_1\xi_3}{(\xi_1+\xi_2)(\xi_3+\xi_2)}}\\
\sqrt{\frac{\xi_2\xi_3}{(\xi_2+\xi_1)(\xi_3+\xi_1)}} & 1 & \sqrt{\frac{\xi_1\xi_2}{(\xi_1+\xi_3)(\xi_2+\xi_3)}}\\
-\sqrt{\frac{\xi_1\xi_3}{(\xi_1+\xi_2)(\xi_3+\xi_2)}} & \sqrt{\frac{\xi_1\xi_2}{(\xi_1+\xi_3)(\xi_2+\xi_3)}} & 1
\end{bmatrix}.
\end{equation}
Now suppose it is known that
\begin{equation}
\phi(\boldsymbol{\mu},\boldsymbol{\xi},a) = \int_{-a}^{a}\int_{-a}^{a}\int_{-a}^{a}f_{Y}(y)dy
\end{equation}
where $f_{Y}(y)$ is the PDF of $Y$ and $a$ is some fixed constant. Let $\boldsymbol{\mu} = (\mu_1,\mu_2,\mu_3)$. It is numerically verified that $\phi(\boldsymbol{\mu},\boldsymbol{\xi},a)$ are equal for $\boldsymbol{\mu} = (-\delta,\delta,0)$, $\boldsymbol{\mu} = (-\delta,0,\delta)$, and $\boldsymbol{\mu} = (0,-\delta,\delta)$ when $\xi_1 = \xi_2 = \xi_3 = 1/3$. I need to prove that this is true only when $\xi_1 = \xi_2 = \xi_3 = 1/3$. (I have verified numerically that these CDFs are equal for $\xi_1 = \xi_2 = \xi_3 = 1/3$.)
Best Answer
The joint cdf of a multivariate distribution uniquely determines the distribution. So, if the cdf's of multivariate normal distributions are the same, then their mean vectors must be the same (and their covariance matrices must be the same).
To simplify the notation, let $d:=\delta$ and $b_i:=\xi_i$.
Then your three multivariate normal distributions will be the same if their covariance matrices are the same and, for instance, $$d=0,\ b_1=0,\ b_2=\frac9{34},\ b_3=\frac{25}{34}$$ -- with unequal $b_i$'s, contrary to your desired conclusion.
If we impose the condition $d\ne0$, then your three multivariate normal distributions can never be the same.
Here are the calculations in Mathematica: