If $\begin{vmatrix}1+a&1&1\\1&1+b&1\\1&1&1+c\end{vmatrix}$, where $a, b$ and $c$ are non zero constants, then the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ is

-1

The correct answer is Option (1) → -1 **

The determinant of the matrix

$\begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix}$

Expanding along the first row:

$(1+a)\big[(1+b)(1+c)-1\big] \;-\; 1\big[1(1+c)-1\big] \;+\; 1\big[1\cdot 1 - (1+b)\cdot 1\big] $

So determinant =

$(1+a)(b+c+bc) - c - b$

Setting determinant = 0:

$(1+a)(b+c+bc) = b + c$

$bc + a(b+c) + abc = 0$

$bc(1+a) + a(b+c) = 0$

Divide by $abc$ (nonzero constants):

$\frac{1+a}{a} + \frac{1}{c} + \frac{1}{b} = 0$

Rewrite:

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = -1$

The required value of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ is $-1$.