Answer
$$0$$
Work Step by Step
$$\eqalign{
& \iiint\limits_D {\left( {xy + xz + yz} \right)}dV \cr
& D = \left\{ {\left( {x,y,z} \right): - 1 \leqslant x \leqslant 1,{\text{ }} - 2 \leqslant y \leqslant 2,{\text{ }} - 3 \leqslant z \leqslant 3} \right\} \cr
& {\text{Substituting the region }}D \cr
& = \int_{ - 3}^3 {\int_{ - 2}^2 {\int_{ - 1}^1 {\left( {xy + xz + yz} \right)} dx} dy} dz \cr
& = \int_{ - 3}^3 {\int_{ - 2}^2 {\left[ {\int_{ - 1}^1 {\left( {xy + xz + yz} \right)} dx} \right]} dy} dz \cr
& {\text{Integrate with respect to }}x \cr
& = \int_{ - 3}^3 {\int_{ - 2}^2 {\left[ {\frac{{{x^2}y}}{2} + \frac{{{x^2}z}}{2} + xyz} \right]_{ - 1}^1} dy} dz \cr
& = \int_{ - 3}^3 {\int_{ - 2}^2 {\left[ {\left( {\frac{y}{2} + \frac{z}{2} + yz} \right) - \left( {\frac{{{{\left( { - 1} \right)}^2}y}}{2} + \frac{{{{\left( { - 1} \right)}^2}z}}{2} + \left( { - 1} \right)yz} \right)} \right]} dy} dz \cr
& = \int_{ - 3}^3 {\int_{ - 2}^2 {\left[ {\left( {\frac{y}{2} + \frac{z}{2} + yz} \right) - \left( {\frac{y}{2} + \frac{z}{2} - yz} \right)} \right]} dy} dz \cr
& = \int_{ - 3}^3 {\int_{ - 2}^2 {\left( {\frac{y}{2} + \frac{z}{2} + yz - \frac{y}{2} - \frac{z}{2} + yz} \right)} dy} dz \cr
& = \int_{ - 3}^3 {\int_{ - 2}^2 {2yz} dy} dz \cr
& {\text{Integrate with respect to }}y \cr
& = \int_{ - 3}^3 {\left[ {{y^2}z} \right]_{ - 2}^2} dz \cr
& = \int_{ - 3}^3 {\left[ {{{\left( 2 \right)}^2}z - {{\left( { - 2} \right)}^2}z} \right]} dz \cr
& = \int_{ - 3}^3 {\left[ {4z - 4z} \right]} dz \cr
& = 0 \cr} $$
