Let a, b, c∈R not all are equal and Δ1=abcbcacab, Δ2=a+2bb+3cc+4ab+2cc+3aa+4bc+2aa+3bb+4c then Δ2Δ1=_________ .

Δ2=ab+3cc+4abc+3aa+4bca+3bb+4c∆'+2bb+3cc+4acc+3aa+4baa+3bb+4c∆''Applying C3→C3−4C1 in Δ' and C2→C2−C1 in ∆'', we getΔ2=ab+3ccbc+3aaca+3bb+2×3bcc+4acaa+4bab   b+4cApplying C2→C2−3C3 in Δ' and C3→C3−C2 in Δ'', we get  =abcbcacab+6bc4aca4bab4c=abcbcacab+24abcbcacab =25abcbcacab=25Δ1⇒ Δ2Δ1=25