Answer
$$V = 8$$
Work Step by Step
$$\eqalign{
& {\text{Let }}F\left( {x,y,z} \right) = 2x + 3y + 6z - 12 \cr
& z = \frac{{12 - 2x - 3y}}{6} \cr
& z = 2 - \frac{1}{3}x - \frac{1}{2}y \cr
& {\text{The region }}D{\text{ is given by:}} \cr
& 2x + 3y + 6z - 12 = 0 \cr
& {\text{Let }}z = 0 \cr
& 2x + 3y + 6\left( 0 \right) - 12 = 0 \cr
& 2x + 3y = 12 \cr
& y = - \frac{2}{3}x + 4 \cr
& D = \left\{ {\left( {x,y,z} \right):0 \leqslant x \leqslant 6,{\text{ 0}} \leqslant y \leqslant - \frac{2}{3}x + 4} \right\} \cr
& {\text{The volume of the solid is }} \cr
& V = \int_0^6 {\int_0^{ - \frac{2}{3}x + 4} {\left( {2 - \frac{1}{3}x - \frac{1}{2}y} \right)} dy} dx \cr
& {\text{Integrate with respect to }}y \cr
& V = \int_0^6 {\left[ {2y - \frac{1}{3}xy - \frac{1}{4}{y^2}} \right]_0^{ - \frac{2}{3}x + 4}} dx \cr
& V = \int_0^6 {\left[ {2\left( { - \frac{2}{3}x + 4} \right) - \frac{1}{3}x\left( { - \frac{2}{3}x + 4} \right) - \frac{1}{4}{{\left( { - \frac{2}{3}x + 4} \right)}^2}} \right]} dx \cr
& {\text{Expanding and simplifying}} \cr
& V = \int_0^6 {\left( {\frac{1}{9}{x^2} - \frac{4}{3}x + 4} \right)} dx \cr
& {\text{Integrating}} \cr
& V = \left[ {\frac{1}{{27}}{x^3} - \frac{2}{3}{x^2} + 4x} \right]_0^6 \cr
& {\text{Evaluating}} \cr
& V = \frac{1}{{27}}{\left( 6 \right)^3} - \frac{2}{3}{\left( 6 \right)^2} + 4\left( 6 \right) \cr
& V = 8 \cr} $$
