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Let x1= A[0] , x2= A[2] , x3= A[4] , y1=A[1], y2=A[3], y3=A[5]
Let d(P1,P2) be the Manhattan distance between points P1=(x1,y1) and P2=(x2,y2). Then d(P1,P2)=|x1−x2|+|y1−y2|.
Note that if you are going from P1 to P2 (or to P3) along the shortest path, the Manhattan distance will be decreasing with each move. So we have to find the next cell that is closer to both P2 and P3.
Now note that if they are in the bounding box of cells P2 and P3 then there are no such “next cell”, since d(X,P2)+d(X,P3) is constant and equal to d(P1,P2) if X is in the bounding box.
Let’s look at the total distance spaceship will travel: from one side it’s d(P1,P2)+d(P1,P3). But from the other side it’s 2⋅d(P1,X)+d(X,P2)+d(X,P3).
So,d(P1,X)=(d(P1,P2)+d(P1,P3)−(d(X,P2)+d(X,P3)))/2 .But since d(X,P2)+d(X,P3)=d(P2,P3) whichever X is chosen we can calculate answer as (d(P1,P2)+d(P1,P3)−d(P2,P3))/2+1 where +1 is because we are talking about cells P1, P2 and P3 and not just points.

Time Complexity: $O(1)$

Space Complexity: $O(1)$