Answer
The average value: $\bar f = \frac{{13}}{{24}}$.
Work Step by Step
We have $f\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2}$ and the region ${\cal W}$ bounded by the planes $2y+z=1$, $x=0$, $x=1$, $z=0$, and $y=0$.
We sketch the solid region and choose to project ${\cal W}$ onto the $xz$-plane and obtain the domain ${\cal S}$. From the figure attached, we see that ${\cal S}$ is a square $0 \le x \le 1$ and $0 \le z \le 1$. It is both vertically and horizontally simple region. However, we choose to describe ${\cal S}$ as a vertically simple region. So, the description of ${\cal S}$ is given by
${\cal S} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,0 \le z \le 1} \right\}$
Thus, ${\cal W}$ is a $y$-simple region consisting of all points lying between ${\cal S}$ and the back face which is the plane $2y+z=1$. So, the description of ${\cal W}$:
${\cal W} = \left\{ {\left( {x,y,z} \right)|0 \le x \le 1,0 \le z \le 1,0 \le y \le \frac{1}{2}\left( {1 - z} \right)} \right\}$
We evaluate the volume of ${\cal W}$:
$Volume\left( {\cal W} \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V$
$ = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{z = 0}^1 \mathop \smallint \limits_{y = 0}^{\left( {1 - z} \right)/2} {\rm{d}}y{\rm{d}}z{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{z = 0}^1 \left( {y|_0^{\left( {1 - z} \right)/2}} \right){\rm{d}}z{\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{z = 0}^1 \left( {1 - z} \right){\rm{d}}z{\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^1 \left( {\left( {z - \frac{1}{2}{z^2}} \right)|_0^1} \right){\rm{d}}x$
$ = \frac{1}{4}\mathop \smallint \limits_{x = 0}^1 {\rm{d}}x = \frac{1}{4}$
To find the average value of $f$, we use Eq. (5):
$\bar f = \frac{1}{{Volume\left( {\cal W} \right)}}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \left( {{x^2} + {y^2} + {z^2}} \right){\rm{d}}V$
$ = 4\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{z = 0}^1 \mathop \smallint \limits_{y = 0}^{\left( {1 - z} \right)/2} \left( {{x^2} + {y^2} + {z^2}} \right){\rm{d}}y{\rm{d}}z{\rm{d}}x$
$ = 4\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{z = 0}^1 \left( {\left( {{x^2}y + \frac{1}{3}{y^3} + y{z^2}} \right)|_0^{\left( {1 - z} \right)/2}} \right){\rm{d}}z{\rm{d}}x$
$ = 4\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{z = 0}^1 \left( {\frac{1}{2}{x^2}\left( {1 - z} \right) + \frac{1}{{24}}{{\left( {1 - z} \right)}^3} + \frac{1}{2}{z^2}\left( {1 - z} \right)} \right){\rm{d}}z{\rm{d}}x$
$ = 4\mathop \smallint \limits_{x = 0}^1 \left( {\left( {\frac{1}{2}{x^2}\left( {z - \frac{1}{2}{z^2}} \right) - \frac{1}{{96}}{{\left( {1 - z} \right)}^4} + \frac{1}{6}{z^3} - \frac{1}{8}{z^4}} \right)|_0^1} \right){\rm{d}}x$
$ = 4\mathop \smallint \limits_{x = 0}^1 \left( {\frac{1}{4}{x^2} + \frac{5}{{96}}} \right){\rm{d}}x$
$ = 4\left( {\frac{1}{{12}}{x^3} + \frac{5}{{96}}x} \right)|_0^1$
$ = 4\left( {\frac{1}{{12}} + \frac{5}{{96}}} \right) = \frac{{13}}{{24}}$
Thus, the average value: $\bar f = \frac{{13}}{{24}}$.
