The key insight is that the squared distance (||p - t||^2 = (p_x - t_x)^2 + (p_y - t_y)^2) between a well at point (p) and tree at point (t) can be obtained by treating each of the two axes independently.
We can project the (x)-coordinates of trees (t_1, \ldots, t_N) onto a number line (the (x)-axis), and likewise for the (y)-coordinates onto the (y)-axis. There may multiple trees with a given (x) or (y), so we can precompute their frequencies in arrays (\text{xcnt}[x]) and (\text{ycnt}[y]). The time and space complexities will both be linear on the largest possible values of (x) and (y), which are only up to (3{,}000).
Then, we see that the inconvenience of a well at point (p) can be computed as:
[\begin{aligned} \sum_{i=1}^{N} \lVert p - t_i \rVert^2 &= \sum_{i=1}^{N} [(p_x - A_i)^2 + (p_y - B_i)^2] \ &= \sum_{i=1}^{N} (p_x - A_i)^2 + \sum_{i=1}^{N} (p_y - B_i)^2 \ &= \sum_{x=0}^{3{,}000} \text{xcnt}[x]\cdot(p_x - x)^2 + \sum_{y=0}^{3{,}000} \text{ycnt}[y]\cdot (p_y - y)^2 \end{aligned}]
Computing this directly will take (\mathcal{O}(3{,}000)) steps on each of the (Q = 500{,}000) queries, for a total of ~(1.5) billion steps. This should already be fast enough to pass within the (6) minute time limit. However, we can optimize further by precomputing the answer for all (3{,}000^2) possible coordinates, after which, each query can be answered in (\mathcal{O}(1)).