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(A) Let $f(x, y) = 2x^2+axy+3y^2+bx+cy-3=0$. Since $(2,-1)$ is the point of intersection,the partial derivatives $\frac{\partial f}{\partial x}$ and $

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$11$

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(A) Let $f(x, y) = 2x^2+axy+3y^2+bx+cy-3=0$. Since $(2,-1)$ is the point of intersection,the partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ must vanish at $(2,-1)$.$\frac{\partial f}{\partial x} = 4x+ay+b = 0 \implies 4(2)+a(-1)+b = 0 \implies 8-a+b = 0 \implies a-b=8$....$(i)$$\frac{\partial f}{\partial y} = ax+6y+c = 0 \implies a(2)+6(-1)+c = 0 \implies 2a+c=6$....(ii)Also,the point $(2,-1)$ satisfies the original equation:$2(2)^2+a(2)(-1)+3(-1)^2+b(2)+c(-1)-3 = 0$$8-2a+3+2b-c-3 = 0 \implies -2a+2b-c = -8 \implies 2a-2b+c = 8$....(iii)From $(i)$,$b = a-8$. Substitute into (iii):$2a-2(a-8)+c = 8 \implies 2a-2a+16+c = 8 \implies c = -8$.Substitute $c=-8$ into (ii):$2a-8 = 6 \implies 2a = 14 \implies a = 7$.From $(i)$,$b = 7-8 = -1$.Finally,$3a+2b+c = 3(7)+2(-1)+(-8) = 21-2-8 = 11$.

If $(2,-1)$ is the point of intersection of the pair of lines $2x^2+axy+3y^2+bx+cy-3=0$,then $3a+2b+c=$

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