First we need to find the range of x for which the function is bound by the line y=12. We need the solution to 12=x+27/x. The solution to x^2-12x+27=0 is x=3 and 9. These will provide the limits of integration later. Now the region is defined. Next we need to find the volume of the bound region when rotated about a line parallel to the y axis, x=-2. It helps to imagine what the rotated region looks like. I see it as a circular gutter with a saucer shaped cross-section filled to the brim with water. Effectively, the volume of the rotated region will be equivalent to the volume of water in the gutter.
Consider the geometry. The radius of rotation will be x+2, because we are rotating around the axis x=-2. Consider a hollow cylinder of thickness dx and radius x+2. The volume of the cylinder will be 2(pi)h(x+2)dx where h is the height of the cylinder for the value of x. This height is the difference between the curve and the line y=12 so its value is 12-(x+27/x). Therefore the volume of the cylindrical shell is 2(pi)(x+2)(12-x-27/x)dx. This is the basis of the integral. If we expand this expression we get 2(pi)(12x+24-x^2-2x-27-54/x)dx=2(pi)(10x-x^2-3+54/x)dx. When this is integrated with respect to x between the limits 3 and 9 we get: 2(pi)[5x^2-x^3/3-3x+54ln(x)] for 3<=x<=9. Evaluating this we get 2(pi)(5(81-9)-(243-9)-3(9-3)+54ln(9/3))=2(pi)(360-234-18+54ln(3))=2(pi)(108+54ln(3))