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If $Δ_{1} = |\begin{array}{l} b + c & a - b & a \\ c + a & b - c & b \\ a + b & c - a & c \end{array}|$ and $Δ_{2} = |\begin{array}{l} a & b & c \\ b & c & a \\ c & a & b \end{array}|$ then $∆_{1} - ∆_{2}$ equal

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Correct option is A

Using C1→C1+C3 and C2→C2-C3 in ∆1 we get Δ1=a+b+c-baa+b+c-cba+b+c-acUsing C1→C1+C2-C3 , we get Δ1=c-baa-cbb-ac=-a-bcb-cac-ab=abcbcacab⇒Δ1=Δ2⇒Δ1-Δ2=0

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If Δ1=b+ca-bac+ab-cba+bc-ac and Δ2=abcbcacab then ∆1-∆2 equal

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If $Δ_{1} = |\begin{array}{l} b + c & a - b & a \\ c + a & b - c & b \\ a + b & c - a & c \end{array}|$ and $Δ_{2} = |\begin{array}{l} a & b & c \\ b & c & a \\ c & a & b \end{array}|$ then $∆_{1} - ∆_{2}$ equal