First of all, don't ever write things like $Z = X+Y+N(0,1)$, not in public as on this forum, not on your exams, and not even in your notebook or private diary: in fact, don't even think of $N(0,1)$ as being the same
as a standard normal random variable. Second, disabuse yourself of the
notion that "$X$ and $Y$ are independent iff the correlation coefficient between $X$ and $Y$ is $r=0$." This assertion is not always true,
not even when
$X$ and $Y$ are restricted to be normal random variables. (It is
true if we restrict $X$ and $Y$ to be jointly normal random
variables; the presence of that adjective jointly is crucial here).
Turning to your question, you seem to be starting from three
independent normal random variables $X, Y, W$ and then considering
the three variables $X, Y$, and $Z = X+Y+W$. Since the vector
$(X,Y,Z)$ is
obtained via a linear transformation from the jointly
normal vector $(X,Y,W)$, $X, Y$, and $Z$ also have a
jointly normal distribution. Note, though, that while
(by assumption) $X, Y, W$ are mutually independent random variables,
$X, Y$, and $Z$ are not mutually independent. In fact, both
$X$ and $Y$ have nonzero correlation with $Z$. Specifically,
\begin{align}
\operatorname{cov}(X,Z) &= \operatorname{cov}(X,X+Y+W)
= \operatorname{cov}(X,X) = \operatorname{var}(X)\\
\operatorname{cov}(Y,Z) &= \operatorname{cov}(Y,X+Y+W)
= \operatorname{cov}(Y,Y) = \operatorname{var}(Y)
\end{align}
Also,
$$\mu_Z = \mu_X + \mu_Y + \mu_W, \quad \text{and}
\quad\operatorname{var}(Z) = \operatorname{var}(X) + \operatorname{var}(Y) + \operatorname{var}(W).$$
Turning to the joint distribution of $X$ and $Y$ conditioned
on $Z = z$, note first that this is a bivariate normal
distribution. It is a standard result (see e.g.
Wikipedia) that
the mean vector of this bivariate distribution is
$$\left[\begin{matrix}\hat{\mu}_X\\ \\ \hat{\mu}_Y \end{matrix}\right]
= \left[\begin{matrix}{\mu}_X
+ \left.\left.\frac{\operatorname{var}(X)}{\operatorname{var}(Z)}\right(z - \mu_Z\right)\\
{\mu}_Y + \left.\left.\frac{\operatorname{var}(Y)}{\operatorname{var}(Z)}\right(z - \mu_Z\right)\end{matrix}\right] $$
while the covariance matrix is
$$\left[\begin{matrix}\operatorname{var}(X) & 0\\
0 & \operatorname{var}(Y)\end{matrix}\right] -
\frac{1}{\operatorname{var}(Z)}\left[\begin{matrix}(\operatorname{var}(X))^2 & \operatorname{var}(X)\operatorname{var}(Y)\\
\operatorname{var}(X)\operatorname{var}(Y) & (\operatorname{var}(Y))^2\end{matrix}\right].$$
Notice that the covariance matrix is not a diagonal matrix,
that is,
$X$ and $Y$ are independent normal random variables (and thus
jointly normal random variables too), but,
conditioned on the value of $Z = X+Y+W$, they are
conditionally dependent (jointly normal) random variables.
Why is this so? Well, broadly speaking,
knowing the value of $Z$ tells us a little
bit about the values of $X$ and $Y$, and what is still unknown
about $X$ and $Y$ is thus linked through this shared information.
Instead of being independent, they are conditionally dependent.
Think of $Z$ as a noisy observation of $X+Y$ where the "noise"
is $W$. If the observation of $X+Y$were completely noiseless, then it
should be obvious that $X$ and $Y$, even though they are independent,
cannot possibly be conditionally independent given that $X+Y = z$ because $Y$ must necessarily equal $z-X$, that is, the value of $X$
determines the value of $Y$. Even for noisy observations, we should
be skeptical of assertions or insistences that $X$ and $Y$ are conditionally independent given that $X+Y+W = z$, and indeed
the math shows that $X$ and $Y$ are conditionally dependent
random variables given the value of a noisy observation $X+Y+W$
of their sum.
