Δ=|a1a2a3b1b2b3c1c2c3|=(a1b2c3+a2b3c1+a3b1c2)−(a3b2c1+a2b1c3+a1b3c2)
Now if x≤3 and y≥-1 then Δ can be maximum 6
But it is not possible as x=3 ⇒each term of x=1 and y=3 ⇒each term of y=-1 ⇒∑3i=1aibici=1 and ∑3i=1aibici=-1 which is contradiction
So now next possibility is 4 which is obtained as
|111-1111-11|=1(1+1−(−1))−(−1)(−1−1−1)=4.