This question can have multiple answers depending on who is doing the answering, or, depending on the interpretation placed on it,
it might be unanswerable.
As I understand the question, we are given that $X_1, X_2$, and $X_3$ are pairwise jointly normal random variables with specified mean vectors
and covariance matrices. We are asked for a trivariate joint normal
distribution that is "most consistent" with this information.
As an engineer (and wannabe probabilist), the first thing I would do
is to check the given information for consistency.
It must be the case that $\mu_{ij} = (\mu_i, \mu_j)$
where $\mu_i = E[X_i]$, $i = 1,2,3$. So, if the problem claims that $\mu_{12} = (4,10)$ while $\mu_{23} = (5,-3)$, the given information
is inconsistent.
Similarly, the covariance matrices must be consistent: in
$\Sigma_{ij}$, the diagonal terms are
$\sigma_i^2 = \operatorname{var}(X_i)$ and
$\sigma_j^2 = \operatorname{var}(X_j)$. Each variance occurs
in two of the three covariance matrices and it must have the same
value in both occurencse. It must also be true that the
off-diagonal terms in $\Sigma_{ij}$ must be equal, and that
$\Sigma_{ij}$ must have determinant $\geq 0$.
If the above consistency checks are satisfied, then, regardless
of whether the $X_i$ are pairwise jointly normal or not, the
mean vector of $(X_1,X_2,X_3)$ is
$$\mu = (\mu_1,\mu_2,\mu_3)\tag{1}$$ and
the covariance matrix is
$$\Sigma = \left[\begin{matrix}
\sigma_1^2& \sigma_{12}& \sigma_{13}\\
\sigma_{21} & \sigma_2^2 & \sigma_{23}\\
\sigma_{31} & \sigma_{32} & \sigma_3^2\end{matrix}\right]\tag{2}$$
in which the upper left $2\times 2$ submatrix is the given
$\Sigma_{12}$, the lower right $2\times 2$ submatrix is the given
$\Sigma_{23}$, and the "four-corners" submatrix is the given $\Sigma_{13}$.
At this point, the probabilist's answer to the question
Which trivariate joint normal distribution is most consistent
with the given information?
would be that $\mathcal N(\mu,\Sigma)$ is the only
trivariate normal distribution that is exactly
consistent with the
given information (that the marginal bivariate distributions
are the specified bivariate joint normal distributions). No other trivariate
normal distribution can be as good a match with the given
information.
On the other hand, the question might be interpreted as
Which trivariate joint normal distribution is the "best
approximation" to the actual unspecified joint distribution
of $(X_1,X_2,X_3)$?
and to my lowly engineering mindset, this question is unanswerable
since neither the meaning of "best" nor the target to be
matched is known;
dyed-in-the-wool probabilists might choose to give a different
response. Other readers who disagree with my
interpretation might want to try their hand at
finding the trivariate joint normal distribution that best approximates
$$f(x_1,x_2,x_3) = \begin{cases}
2\phi(x_1)\phi(x_2)\phi(x_3),
& \text{for}~ x_1 \geq 0, x_2 \geq 0, x_3 \geq 0,\\
&\text{ or}~ x_1<0, x_2< 0, x_3 \geq 0,\\
&\text{ or}~ x_1<0, x_2\geq 0, x_3 < 0,\\
&\text{ or}~ x_1\geq 0, x_2< 0, x_3 < 0,\\
0, &\text{otherwise},\end{cases}$$
where $\phi(\cdot)$ is the standard normal density function.
Note that the $X_i$ are pairwise independent standard normal
random variables.
On the other hand, a good statistician might take it that the OP
has masses of data (maybe even "Big Data") from which
the OP has drawn the inference that it is reasonable to
assume that the $X_i$ are pairwise jointly normal
random variables, and that $\mu_{12}$, $\mu_{23}$, $\mu_{13}$, $\Sigma_{12}$
$\Sigma_{23}$, $\Sigma_{13}$ are sample means, variances, and covariances
gleaned from three different data sets
of the form $\{(x_{1,i}, x_{2,i})\colon i = 1,2,\ldots n_1\}$,
$\{(x_{2,j},x_{3,j})\colon j = 1,2,\ldots, n_2\}$ and
$\{(x_{1,k},x_{3,k})\colon k = 1,2,\ldots n_3\}$. Then it
is almost certain that the first entry in $\mu_{12}$
(which is $\frac{1}{n_1}\sum_i x_{1,i}$) is not the same
as the first entry in $\mu_{13}$ (which is $\frac{1}{n_3}\sum_k x_{1,k}$).
Similarly for the other means, variances and covariances. So,
should the data be merged to get better estimates of what the
means, variances, etc are? Is there data available of the
form $\{(x_{1,m}, x_{2,m}, x_{3,m})\colon m = 1,2,\ldots N\}$
that might allow some form of estimation as to what the
actual joint distribution of $(X_1,X_2,X_3)$ is?
Considerations of this type
lead to the kinds of answers given by Tim and Karel Macek.
Best Answer
Yes, the two probabilities ought to be different, because one is for a mixture and the other is for a sum. Look at an example:
The thick red curve is the probability density function for a mixture of three normals ($X$). The dashed curves are its components (each scaled by $\lambda_i$); they are normal. The thick blue curve is the pdf of the normal distribution with the weighted mean and weighted variance that define $Y$; it, too, is normal. In particular, note that the possibility of the mixture having multiple modes (three in this case, between one and three in general) makes it perfectly clear the mixture is not normal in general, because normal distributions are unimodal.
The mixture can be modeled as a two step process: first draw one of the three ordered pairs $(\mu_1, \sigma_1)$, $(\mu_2, \sigma_2)$, and $(\mu_3, \sigma_3)$ with probabilities $\lambda_1$, $\lambda_2$, and $\lambda_3$, respectively. Then draw a value $X$ from the normal distribution specified by the parameters you drew (understood as mean and standard deviation).
The weighted mean is obtained from a completely different procedure: independently draw a value $X_1$ from a normal distribution with parameters $(\lambda_1 \mu_1, \lambda_1 \sigma_1)$, a value $X_2$ from a normal distribution with parameters $(\lambda_2 \mu_2, \lambda_2 \sigma_2)$, and a value $X_3$ from a normal distribution with parameters $(\lambda_3 \mu_3, \lambda_3 \sigma_3)$. Then form their sum $Y = X_1+X_2+X_3$.