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If Δ=∣1+a1111+b1111+c∣\Delta = \begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix}, Then

Match List-I with List-II

List-IList-II
Values of a,ba, b & ccvalue of Δ\Delta
(A). a=b=c=1a=b=c=1(I). 2
(B). a=−1,b=1,c=1a=-1,b=1,c=1(II). 0
(C). a=−1,b=−1,c=1a=-1,b=-1,c=1(III). -2
(D). a=−1,b=−1,c=−1a=-1,b=-1,c=-1(IV). 4

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 4

Expanding, Δ=abc+ab+bc+ca\Delta=abc+ab+bc+ca. For a=b=c=1a=b=c=1: 1+3=41+3=4 (IV). For a=−1,b=c=1a=-1,b=c=1: −1−1=−2-1-1=-2 (III). For a=b=−1,c=1a=b=-1,c=1: 1−1=01-1=0 (II). For a=b=c=−1a=b=c=-1: −1+3=2-1+3=2 (I).

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