**Hint 1**: Observe that the XOR of a subarray from `i` to `j` can be computed efficiently using a prefix XOR array. Precompute `prefix_xor` where `prefix_xor[i]` is the XOR of all elements from `arr[0]` to `arr[i-1]`. Then, the XOR of any subarray `arr[i..j]` is `prefix_xor[j+1] ^ prefix_xor[i]`.

**Hint 2**: For a triplet `(i, j, k)` to satisfy the condition, the XOR of `arr[i..j-1]` must equal the XOR of `arr[j..k]`. Using the prefix XOR array, this translates to `prefix_xor[j] ^ prefix_xor[i] == prefix_xor[k+1] ^ prefix_xor[j]`. Simplify this to `prefix_xor[i] == prefix_xor[k+1]`.

**Hint 3**: The problem now reduces to counting the number of triplets `(i, j, k)` where `i < j <= k` and `prefix_xor[i] == prefix_xor[k+1]`. For each `k`, iterate over all possible `i` (where `prefix_xor[i] == prefix_xor[k+1]` and `i < k+1`) and count the number of valid `j` values (where `i < j <= k`). Optimize this by precomputing the frequency of each `prefix_xor` value and using it to count valid `i` values efficiently.