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If a, b and c are all different from zero and ∣1+a1111+b1111+c∣=0\begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} = 0, then the value of 1a+1b+1c\frac{1}{a} + \frac{1}{b} + \frac{1}{c} is :

Solution

✅ Correct Option: 3

Taking out aa, bb, cc from rows after dividing each row appropriately, the determinant simplifies to abc(1+1a+1b+1c)=0abc\left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) = 0. Since a,b,c≠0a, b, c \neq 0: 1a+1b+1c=−1\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = -1.

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