Conditional Expectation / Estimator Confusion Let $X_1, X_2, X_3 \sim N(0, d^2)$ and $T = X_1^2 + X_2^2 + X_3^2.$ 
I have an estimator for $d$, $$\hat{d} = \frac{\sqrt{T\ 2\pi}}{4},$$
and another estimator for $d$, $$\tilde{d} = \frac{1}{3} \sqrt{\pi / 2}\ \left(|X_1| + |X_2| + |X_3|\right).$$
I need to show that $\mathbb{E}(\tilde{d} | T) = \hat{d}$ and that $\mathbb{E}(|X_1| | T) = \frac{1}{2} \sqrt{T}$.
I'm not clear how conditioning the expectation on $T$ changes things and how it affects the computation of expectation.  

(The following edit was attempted by the OP in an answer, I'm updating the question on their behalf.) 
I'm mostly confused about how to write out $E[d\sim | T]$ or $E[|X1| | T]$ and begin to work through the algebra.
 A: Since this looks like homework, or exercise I will skim through the solution.
Let's write
$$f_{X_i|T}(x_i | t) = \frac{f_{X_i, T}(x_i, t)}{f_T(t)}.$$
The joint distribution $f_{X_i, T}(x_i, t)$ is the product of a Gaussian $(0, d^2)$ and a $\chi^2(2)$ (the post does not specify that $X_1, X_2, X_3$ are independent, but I will take it for granted):
$$f_{X_i, T}(x_i, t) \propto \exp\left\{-\frac{x_i^2}{2d^2} \right\} \exp\left\{-\frac{t-x_i^2}{2d^2}\right\},$$
which (miraculously) simplifies to
$$f_{X_i, T}(x_i, t) \propto \exp\left\{-\frac{t}{2d^2}\right\}.$$
From here this is quite easy because we realize that the numerator does not depend on $x_i$ so the distribution of $x_i$ given $t$ is uniform. The boundaries for $x_i$ given $t$ are $-\sqrt{t}$ and $+\sqrt{t}$ by construction, so the expected value $E(|X_i| | T=t)$ is 
$$ 2\int_0^{\sqrt{t}} \frac{x_i}{2\sqrt{t}} dx_i = \frac{1}{2}\sqrt{t}.$$
The expected value of a sum is the sum of the expected values, so getting $E(\tilde{d} | T=t)$ is piece of cake.
EDIT: Following @whuber's comment, here is how I got $f_{X_i|T}(x_i | t)$. I will assume $i=1$ for the sake of the argument. Let's define $Z = X_2^2 + X_3^2$, which is a scaled $\chi^2(2)$, i.e with 2 degrees of freedom.
The joint density of $(X_1, Z)$ is the product of their densities because they depend on different variables, so $\propto \exp\left\{-\frac{x_1^2}{2d^2} \right\} \exp\left\{-\frac{z}{2d^2}\right\}$. Now we use the change of variable $(X_1, Z) \rightarrow (X_1, X_1^2 + Z = T)$. The Jacobian of the inverse transformation is 1, so $f_{X_1|T}(x_1 | t) \propto \exp\left\{-\frac{x_1^2}{2d^2} \right\} \exp\left\{-\frac{t-x_1^2}{2d^2}\right\}$.
