Discussion

Explanation:

From (1), X, U, Z are at b,c, g or at b, f, g in some order. Thus, X, U or Z is definitely at g.

Let X is at g.

Case (i): U and Z at b and f.

From (4), U has to be at b, Z at f and V at j. From (2), no one is at c, e, k and h. As Y sees both U and W, Y must be at a and W at i. But then W sees V, which contradicts (5).
Thus, this case is not valid.

Case (ii): U and Z at b and c.

From (4), U has to be at c, Z at b and V at a. From (2), no one is at c, e,f, k and h. But then Y must be at I, j or l. But in that case Y cannot see U. Thus, this case is not valid.

Therefore, X cannot be at g.

Let Z is at g.

From (4), U is at c or f.

Case (i) U is at c and hence V is at k and X is at a. Again there is no place for V. This case is invalid.

Case (ii) U is at f, X is at b and V at h. V will be at e as he sees U. But then he will be able to see Z also. So this case is also invalid.

Thus, U is at g. Therefore, Z is at f, V at e and X at b. So, from (2) no one will be at a, c and j. From (3) and (5), it can be concluded that Y is at k and W at l. Thus, we have

No one is standing at intersection a.

Hence, option (a).

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