Answer
$$48\ln 6 - 180$$
Work Step by Step
$$\eqalign{
& \int_1^6 {\int_0^{4 - 2y/3} {\int_0^{12 - 2y - 3z} {\frac{1}{y}dxdzdy} } } \cr
& {\text{Integrate with respecto to }}x \cr
& = \int_1^6 {\int_0^{4 - 2y/3} {\left[ {\frac{x}{y}} \right]_0^{12 - 2y - 3z}dzdy} } \cr
& = \int_1^6 {\int_0^{4 - 2y/3} {\left[ {\frac{{12 - 2y - 3z}}{y}} \right]dzdy} } \cr
& = \int_1^6 {\int_0^{4 - 2y/3} {\left( {\frac{{12}}{y} - 2 - \frac{{3z}}{y}} \right)dzdy} } \cr
& {\text{Integrate with respecto to }}z \cr
& = \int_1^6 {\left[ {\frac{{12z}}{y} - 2z - \frac{{3{z^2}}}{2}} \right]_0^{4 - 2y/3}} dy \cr
& = \int_1^6 {\left[ {\frac{{12}}{y}\left( {4 - \frac{{2y}}{3}} \right) - 2\left( {4 - \frac{{2y}}{3}} \right) - \frac{3}{2}{{\left( {4 - \frac{{2y}}{3}} \right)}^2}} \right]} dy \cr
& = \int_1^6 {\left[ {\frac{{48}}{y} - 8 - 8 + \frac{{4y}}{3} - \frac{3}{2}\left( {16 - \frac{{16y}}{3} + \frac{{4{y^2}}}{9}} \right)} \right]} dy \cr
& = \int_1^6 {\left( {\frac{{48}}{y} - 16 + \frac{{4y}}{3} - 24 + 8y - \frac{{2{y^2}}}{3}} \right)} dy \cr
& = \int_1^6 {\left( {\frac{{48}}{y} + \frac{{28y}}{3} - 40 - \frac{{2{y^2}}}{3}} \right)} dy \cr
& {\text{Integrate and evaluate}} \cr
& = \left[ {48\ln \left| y \right| + \frac{{14{y^2}}}{3} - 40y - \frac{{2{y^3}}}{3}} \right]_1^6 \cr
& = \left[ {48\ln \left| 6 \right| + \frac{{14{{\left( 6 \right)}^2}}}{3} - 40\left( 6 \right) - \frac{{2{{\left( 6 \right)}^3}}}{3}} \right] - \left[ {\frac{{14{{\left( 1 \right)}^2}}}{3} - 40\left( 1 \right) - \frac{{2{{\left( 1 \right)}^3}}}{3}} \right] \cr
& = 48\ln 6 - 216 + 36 \cr
& = 48\ln 6 - 180 \cr} $$
