In order for [tex]f(x,y,z)[/tex] to be a proper density function, we need to have
[tex]\displaystyle\int_{\mathcal S}f(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=1[/tex]
where [tex]\mathcal S[/tex] is the support of the joint distribution. Here [tex]\mathcal S[/tex] is the box with dimensions 1x2x2. We have
[tex]\displaystyle\int_{\mathcal S}f(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=c\int_{x=0}^{x=1}x\,\mathrm dx\int_{y=0}^{y=2}y\,\mathrm dy\int_{z=0}^{z=2}z\,\mathrm dz[/tex]
[tex]=\dfrac{2^2c}2=1\implies c=\dfrac12[/tex]
Now since we can write [tex]f_{X,Y,Z}(x,y,z)[/tex] as
[tex]\dfrac{xyz}2=\dfrac x{c_x}\cdot\dfrac y{c_y}\cdot\dfrac z{c_z}[/tex]
(where [tex]c_xc_yc_z=2[/tex]) we can assign each factor to be the PDF for the corresponding random variable, i.e.
[tex]f_{X,Y,Z}(x,y,z)=f_X(x)\cdot f_Y(y)\cdot f_Z(z)[/tex]
which means that [tex]X,Y,Z[/tex] are independent of one another. Then
[tex]P(X\le1,Y\le1,Z\le1)=P(X\le1)\cdot P(Y\le1)\cdot P(Z\le1)[/tex]
We can derive the marginal distributions in the same way we solve for [tex]c[/tex] above:
[tex]\displaystyle\int_{x=0}^{x=1}\frac x{c_x}\,\mathrm dx=1\implies c_x=\frac12[/tex]
[tex]\implies f_X(x)=\begin{cases}2x&\text{for }0\le x\le1\\0&\text{otherwise}\end{cases}[/tex]
[tex]\displaystyle\int_{y=0}^{y=2}\frac y{c_y}\,\mathrm dz=\int_{z=0}^{z=2}\frac z{c_z}\,\mathrm dz=1\implies c_y=c_z=2[/tex]
[tex]\implies f_Y(y)=\begin{cases}\dfrac y2&\text{for }0\le y\le2\\\\0&\text{otherwise}\end{cases}[/tex]
with [tex]f_Z(z)[/tex] defined similarly, so that
[tex]P(X\le 1,Y\le1,Z\le1)=1\cdot\dfrac14\cdot\dfrac14=\dfrac18[/tex]
For the final probability, we condition [tex]Z[/tex] on the event that [tex]X=x[/tex] and [tex]Y=y[/tex]:
[tex]P(X+Y+Z\le1)=P(Z\le1-X-Y)[/tex]
[tex]=\displaystyle\iint_{(x,y)}P(Z<1-X-Y\mid X=x,Y=y)\cdot P(X=x,Y=y)\,\mathrm dx\,\mathrm dy[/tex]
[tex]=\displaystyle\int_{x=0}^{x=1}\int_{y=0}^{y=2}F_Z(1-x-y)f_X(x)f_Y(y)\,\mathrm dy\,\mathrm dx[/tex]
The CDF of [tex]Z[/tex] is easy to find:
[tex]F_Z(z)=\begin{cases}0&\text{for }z<0\\\\\dfrac{z^2}4&\text{for }0\le z<2\\\\1&\text{for }z\ge2\end{cases}[/tex]
So we have
[tex]P(X+Y+Z\le1)=\displaystyle\int_{x=0}^{x=1}\int_{y=0}^{y=2}\frac{(1-x-y)^2}4\cdot2x\cdot\frac y2\,\mathrm dx\,\mathrm dy=\frac{23}{72}[/tex]
