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If a,ba, b and cc are non-zero real number such that ∣1+a1111+b1111+c∣=0\begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} = 0, then value of 1a+1b+1c\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} is

Solution

✅ Correct Option: 3

Taking aa, bb, cc common from the rows after dividing and using row operations, the determinant reduces to abc(1+1a+1b+1c)abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right). Since abc≠0abc \neq 0 and the determinant is 00, 1a+1b+1c=−1\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = -1.

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