Given that $b^2-ac < 0, a < 0, a >0. $ The value of

$Δ=\begin{bmatrix}a & b & ax+by\\b & c & bx+cy\\ax+by & bx+cy & 0\end{bmatrix}, $ is

negative

The correct answer is option (3) : negative

$Δ=\begin{bmatrix}a & b & ax+by\\b & c & bx+cy\\ax+by & bx+cy & 0\end{bmatrix}$

$⇒Δ=\begin{bmatrix}a & b & ax+by\\b & c & bx+cy\\0& 0 & -(ax^2+2bxy+cy^2)\end{bmatrix}$           $\begin{bmatrix}Applying \, R_3→ \\ R_3-xR_1-yR_2\end{bmatrix}$

$⇒Δ= (b^2 -ac) (ax^2 +2bxy + cy^2)$

Now,

$b^2-ac< 0 $ and $a > 0 $

⇒ Discriminant of $ax^2 + 2bxy + cy^2 $ is negative and a > 0.

$⇒ax^2 + 2bxy + cy^2 > 0 \, ∀ \, x, y \in R.$

$⇒Δ=(b^2-ac) (ax^2 +2bxy +cy^2) < 0$