Step 1: Recall the key identity.
The distributive law of Boolean algebra gives $(A+B)(A+C)=A+BC$. We will reuse this twice.
Step 2: Group the first two factors.
In $(x+y'+z')(x+y'+z)$, treat $A=x+y'$. The two factors become $(A+z')(A+z)$.
Step 3: Collapse them.
Using the identity with $B=z'$ and $C=z$, $(A+z')(A+z)=A+z'z=A+0=A$. So the first two factors reduce to $x+y'$.
Step 4: Bring in the third factor.
Now multiply $(x+y')(x+y+z')$. Take $A=x$, $B=y'$, $C=y+z'$, so the product is $x+y'(y+z')$.
Step 5: Expand the inner product.
$y'(y+z')=y'y+y'z'=0+y'z'=y'z'$.
Step 6: Write the final form.
Hence the expression simplifies to $x+y'z'$.
\[ \boxed{x+y'z'} \]