Let $(R,\mathfrak m)$ be a regular local ring. Let $I,J$ be proper ideals of $R$ such that $R/(I+J)$ has finite length i.e. $\sqrt{I+J}=\mathfrak m.$ Since $I+J$ annihilates $\text{Tor}_n^R(R/I, R/J)$ for each $n$, so each $\text{Tor}_n^R(R/I, R/J)$ has finite length. Consider Serre's intersection multiplicity $$\chi(R/I, R/J):=\sum_{n=0}^\infty (-1)^n \operatorname{length}_R\text{Tor}_n^R(R/I, R/J)$$

Now, also assume $\dim(R/I)+\dim(R/J)=\dim R$. (i.e. $V(I)$ and $V(J)$ intersect properly).

My question is: If $\chi(R/I, R/J)=\operatorname{length}_R (R/(I+J))$, then is it true that $\text{Tor}_i^R(R/I, R/J)=0$ for all $i\ge 1$ ? (or equivalently, are both $R/I$ and $R/J$ Cohen-Macaulay?)

If needed, I am willing to assume $I,J$ are prime ideals.

Note: Here is an argument that why the vanishing of all positive Tor is equivalent to saying $R/I, R/J$ are Cohen-Macaulay: Indeed, since $R$ is regular and each $\text{Tor}_n^R(R/I, R/J)$ has depth $0$ (since finite length), so putting $q:=\sup\{n: \text{Tor}_n^R(R/I, R/J)\ne 0 \}$, we see by Theorem 2.2 of https://doi.org/10.1080/00927879808826375 that $$q=\text{depth }R- \text{depth }(R/I) - \text{depth }(R/J)=\dim(R/I)+\dim(R/J)- \text{depth }(R/I) - \text{depth }(R/J),$$ where we used the assumption that $V(I)$ and $V(J)$ intersect properly i.e. $\dim(R/I)+\dim(R/J)=\dim R$. Hence, $q=0$ if and only if $\dim(R/I)- \text{depth }(R/I) +\dim(R/J) - \text{depth }(R/J)=0$ if and only if $\dim(R/I)- \text{depth }(R/I) =0=\dim(R/J) - \text{depth }(R/J)$ i.e. both $R/I$ and $R/J$ are Cohen-Macaulay.