Answer
$$V = 12$$
Work Step by Step
$$\eqalign{
& 2x + 3y + 4z = 12 \cr
& 4z = 12 - 2x - 3y \cr
& z = 3 - \frac{1}{2}x - \frac{3}{4}y \cr
& {\text{The volume of the solid is given by}} \cr
& V = \iint\limits_R zdA,{\text{ }}z = 3 - \frac{1}{2}x - \frac{3}{4}y \cr
& 2x + 3y + 4\left( 0 \right) = 12 \cr
& 2x + 3y = 12 \cr
& y = 4 - \frac{2}{3}x \cr
& {\text{Let the region be}} \cr
& R = \left\{ {\left( {x,y} \right):0 \leqslant y \leqslant 4 - \frac{2}{3}x,{\text{ }}0 \leqslant x \leqslant 6} \right\} \cr
& V = \iint\limits_R zdA = \int_0^6 {\int_0^{4 - \frac{2}{3}x} {\left( {3 - \frac{1}{2}x - \frac{3}{4}y} \right)} } dydx \cr
& {\text{Integrate with respect to }}y \cr
& V = \int_0^6 {\left[ {3y - \frac{1}{2}xy - \frac{3}{8}{y^2}} \right]_0^{4 - \frac{2}{3}x}} dx \cr
& V = \int_0^6 {\left[ {3\left( {4 - \frac{2}{3}x} \right) - \frac{1}{2}x\left( {4 - \frac{2}{3}x} \right) - \frac{3}{8}{{\left( {4 - \frac{2}{3}x} \right)}^2}} \right]} dx \cr
& V = \int_0^6 {\left( {12 - 2x - 2x + \frac{1}{3}{x^2} - 6 + 2x - \frac{1}{6}{x^2}} \right)} dx \cr
& V = \int_0^6 {\left( {6 - 2x + \frac{1}{6}{x^2}} \right)} dx \cr
& {\text{Integrate}} \cr
& V = \left[ {6x - {x^2} + \frac{1}{{18}}{x^3}} \right]_0^6 \cr
& V = \left[ {6\left( 6 \right) - {{\left( 6 \right)}^2} + \frac{1}{{18}}{{\left( 6 \right)}^3}} \right] - \left[ 0 \right] \cr
& V = 12 \cr} $$
