If for some c < 0, the quadratic equation $2cx^2 -2(2c-1) x+3c^2 =0$ has two distinct real roots $\frac{1}{a}$ and $\frac{1}{b}$, then the value of the determinant $\begin{vmatrix} 1+a & 1 & 1\\1 & 1+b & 1\\ 1 & 1 & 1+c\end{vmatrix},$ is

2

The correct answer is option (1) : 2

It is given that $\frac{1}{a}$ and $\frac{1}{b}$ are roots of the given equation.

$∴\frac{1}{a}+\frac{1}{b}=\frac{2(2c-1)}{2c}$ and $\frac{1}{a}×\frac{1}{b}=\frac{3c^2}{2c}$

$⇒\frac{1}{a}+\frac{1}{b}=2-\frac{1}{c}$ and $\frac{1}{b}=\frac{3c}{2}$

$⇒\frac{1}{a}+\frac{1}{b}+\frac{1}{c}= 2 $ and $\frac{1}{abc}= \frac{3}{2}$

$⇒\frac{1}{a}+\frac{1}{b}+\frac{1}{c}= 2 $ and $abc=\frac{2}{3}$

Now, $\begin{vmatrix} 1+a & 1 & 1\\1 & 1+b & 1\\ 1 & 1 & 1+c\end{vmatrix}=abc \left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$

$\begin{vmatrix} 1+a & 1 & 1\\1 & 1+b & 1\\ 1 & 1 & 1+c\end{vmatrix}=\frac{2}{3}(1+2) = 2 $