How to decrypt unusual Many Time Pad How does one decrypt the many time pad:

($R$ is a random string, $C$ is ciphertext, $P$ is plaintext)

$R_0 \oplus R_1 = C_0$

$R_2 \oplus R_1 = C_1$

$P_0 \oplus R_1 = C_2$

assuming that $R_n$ are perfectly random pads, plaintext is writing and $C_n$ are the encrypted pads. The key $R_1$ is used three times but it doesn't seem like Crib Dragging (the solution I've seen described to decrypt many time pads) would work (edit: to retrieve $R_0$, $R_2$, and $P_0$ or even just $P_0$) since the other values that $R_1$ is added to are random. Is there something I'm missing?
This cannot be used to compromise $P_0$. Suppose that, instead of generating $C_0$ and $C_1$ as $R_0\oplus R_1$ and $R_2\oplus R_1$, the attacker instead picked $C_0$ and $C_1$ at random from the set of bit strings the length of $C_2$. This situation is indistinguishable from yours; in both cases, $R_0$ and $R_2$ are independent and follow a uniform probability distribution, and the same is true for $C_0$ and $C_1$. If you can compromise $P_0$ in your case, you can compromise it if the sender had actually generated $C_0$ and $C_1$. But that would mean that sending two completely random strings alongside an OTP message somehow helps compromise the message. Obviously, that's not true; the attacker could otherwise create his own random strings and break the OTP that way.