\begin{aligned} &\text{Let, } \phi(x, y) = x^2 + 3xy + 2y^2 - x - 4y - 6 = 0 \\ &\text{Taking the partial derivative w.r.t. } x, \text{ we get:} \\ &\frac{\partial \phi}{\partial x} = 2x + 3y - 1 = 0 \quad \cdots (1) \\ &\text{Taking the partial derivative w.r.t. } y, \text{ we get:} \\ &\frac{\partial \phi}{\partial y} = 3x + 4y - 4 = 0 \quad \cdots (2) \\ &\text{Multiplying Equation (1) by 3 and Equation (2) by 2, we get:} \\ &6x + 9y - 3 = 0 \quad \cdots (3) \\ &6x + 8y - 8 = 0 \quad \cdots (4) \\ &\text{Now, subtracting Equation (4) from Equation (3), we get:} \\ &(6x + 9y - 3) - (6x + 8y - 8) = 0 \\ &y + 5 = 0 \Rightarrow y = -5 \\ &\text{From Equation (1), substituting } y = -5: \\ &2x + 3(-5) - 1 = 0 \\ &2x - 15 - 1 = 0 \\ &2x = 16 \\ &x = 8 \\ &\text{So, the point of intersection is } (8, -5). \\ &\text{Hence, option (3) is correct.} \end{aligned}