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(D) The equation of the hyperbola is $\frac{(x-3)^2}{3}-\frac{(y-2)^2}{2}=1$. Expanding this,we get $2(x^2-6x+9) - 3(y^2-4y+4) = 6$,which simplifies t

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$2 x^2-3 y^2-12 x+12 y+6=0$

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(D) The equation of the hyperbola is $\frac{(x-3)^2}{3}-\frac{(y-2)^2}{2}=1$. Expanding this,we get $2(x^2-6x+9) - 3(y^2-4y+4) = 6$,which simplifies to $2x^2 - 3y^2 - 12x + 12y + 18 - 12 = 6$,or $2x^2 - 3y^2 - 12x + 12y = 0$. The equation of the pair of asymptotes differs from the hyperbola equation only by a constant term. Let the equation of the pair of asymptotes be $2x^2 - 3y^2 - 12x + 12y + \lambda = 0$. For this to represent a pair of straight lines,the determinant $\Delta$ must be zero. Comparing with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we have $a=2, b=-3, h=0, g=-6, f=6, c=\lambda$. The condition $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$ gives $2(-3)(\lambda) + 2(6)(-6)(0) - 2(6)^2 - (-3)(-6)^2 - \lambda(0)^2 = 0$. This simplifies to $-6\lambda - 72 + 108 = 0$,which gives $-6\lambda + 36 = 0$,so $\lambda = 6$. Thus,the equation of the pair of asymptotes is $2x^2 - 3y^2 - 12x + 12y + 6 = 0$.

The equation of the pair of asymptotes of the hyperbola $\frac{(x-3)^2}{3}-\frac{(y-2)^2}{2}=1$ is

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