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Consider a determinant Δ=∣3a−a+b−a+c−b+a3b−b+c−c+a−c+b3c∣\Delta=\begin{vmatrix}3a & -a+b & -a+c\\ -b+a & 3b & -b+c\\ -c+a & -c+b & 3c\end{vmatrix}, then

Match the LIST-I with LIST-II

LIST-I (Value of a,b,ca, b, c)LIST-II (Value of Δ\Delta)
A.a=0,b=0,c=1a = 0, b = 0, c = 1I.2727
B.a=0,b=1,c=1a = 0, b = 1, c = 1II.00
C.a=0,b=2,c=−1a = 0, b = 2, c = -1III.66
D.a=1,b=1,c=1a = 1, b = 1, c = 1IV.−6-6

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 2

Substitute and expand each case.

A: the first two rows both become (0,0,1)(0,0,1), so Δ=0\Delta=0 (II).

B: rows (0,1,1),(−1,3,0),(−1,0,3)(0,1,1),(-1,3,0),(-1,0,3) give Δ=6\Delta=6 (III).

C: rows (0,2,−1),(−2,6,−3),(1,3,−3)(0,2,-1),(-2,6,-3),(1,3,-3) give Δ=−6\Delta=-6 (IV).

D: the matrix is 3I3I, so Δ=27\Delta=27 (I).

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