First, sort $$$E_{i\cdot Y + j} A_i + B_j$$$, which has $$$X \times Y$$$ elements, in decreasing order. Then resize this long array $$$E$$$ to first $$$K$$$ value. Then, get first $$$K$$$ elements of $$$E_i + C_j$$$ by sorting.
The time complexity is $$$O(XY \log XY + KZ \log KZ)$$$.

First, sort array $$$A, B, C$$$ into descending order.
Then, the only candidate of top-$$$K$$$ value is $$$A_i + B_j + C_k$$$ where $$$i \times j \times k \leq K$$$.
The number of candidates is 106,307. Hence, we can get top-$$$K$$$ value by sorting, which fits into the time limit.
In fact, the number of candidates is $$$O(K \log^2 K)$$$ for general $$$K$$$. Thus, the time complexity is $$$O(K \log^3 K)$$$.

First, sort $$$A, B, C$$$ into descending order.
The main observation is that, if $$$(A_i, B_j, C_k)$$$ is not selected, then any of $$$(A_{i+1}, B_j, C_k), (A_i, B_{j + 1}, C_k), (A_i, B_j, C_{k + 1})$$$ will not be selected.
We can start from $$$(A_1, B_1, C_1)$$$ and process with priority queue, pushing three triplets $$$(A_{i+1}, B_j, C_k), (A_i, B_{j + 1}, C_k), (A_i, B_j, C_{k + 1})$$$ when $$$(A_i, B_j, C_k)$$$ is selected. Repeat this process until $$$K$$$ elements will be selected.
The time complexity is $$$O(K \log K)$$$.

Think about binary-searching the answer. Do kind of pruning brute-force using sorted $$$A, B, C$$$. We can judge if the answer is less than $$$K$$$ or not, in time complexity $$$O(K + X + Y + Z)$$$.
The total time complexity is $$$O((K + X + Y + Z) \log max)$$$.