Without loss of generality, assume $X, U, V \text{ i.i.d. } \sim N(0, 1)$ (in general the analysis below requires the joint normality of $(X, U, V)$). The joint normality of $(X, U, V)$ and transformations
\begin{align*}
\begin{bmatrix}
X \\
S \\
T
\end{bmatrix} =
\begin{bmatrix}
1 & 0 & 0 \\
1 & 1 & 0 \\
1 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
X \\
U \\
V
\end{bmatrix}, \quad
\begin{bmatrix}
X \\
R
\end{bmatrix}
= \begin{bmatrix}
1 & 0 & 0 \\
a + b & a & b
\end{bmatrix}
\begin{bmatrix}
X \\
U \\
V
\end{bmatrix}
\end{align*}
then indicate that
\begin{align*}
\begin{bmatrix}
X \\
S \\
T
\end{bmatrix} \sim N_3\left(0,
\begin{bmatrix}
1 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 2
\end{bmatrix}\right),
\quad
\begin{bmatrix}
X \\
R
\end{bmatrix} \sim
N_2\left(0,
\begin{bmatrix}
1 & a + b \\
a + b & 2a^2 + 2ab + 2b^2
\end{bmatrix}\right).
\end{align*}
It then follows by the conditional distribution of multivariate normal distribution that
\begin{align*}
& E[X|S, T] = \begin{bmatrix} 1 & 1 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 1 & 2
\end{bmatrix}^{-1}\begin{bmatrix} S \\ T \end{bmatrix} = \frac{1}{3}(S + T), \\
& E[X|R] = \frac{a + b}{2a^2 + 2ab + 2b^2}R = \frac{a + b}{2a^2 + 2ab + 2b^2}(aS + bT).
\end{align*}
So the equality holds if and only if
\begin{align*}
\begin{cases}
3a(a + b) = 2a^2 + 2ab + 2b^2 \\
3b(a + b) = 2a^2 + 2ab + 2b^2
\end{cases}.
\end{align*}
This equation holds for any $a, b$ such that $a = b$.
For the general $(X, U, V) \sim N(0, \Sigma)$ case (the zero-mean condition is essential and cannot be relaxed), repeating the argument above yields $E[X|S, T] = AS + BT$ for some constants $A, B$ depending on $\Sigma$ (in the above i.i.d. case, $A = B = 1/3$). Now apply the argument in Yves's answer would give you at least one choice of $(a, b) := (A, B)$, because $\sigma(R) = \sigma(AS + BT) \subset \sigma(S, T)$ and the law of iterative expectations:
\begin{align*}
E[X|R] = E[E[X|S, T]|R] = E[R|R] = R = E[X|S, T].
\end{align*}
