Answer
$$0.01$$
Work Step by Step
$f_x=2x+y $ and $ f_x(1,1,2)=2(1)+1=3$
and $f_y =x+z $ and $f_{y}(1,1,2) =1+2=3 \\ f_z=y+\dfrac{z}{2} $ and $f_z(1,1,2)=1+\dfrac{2}{2}=2$
Now $$f_{xx}=2 \\ f_{yy}=0 \\ f_{zz} =\dfrac{1}{2} \\ f_{xy}=1 \\ f_{xx}= 0 \\ f_{yz}=1$$
The error can be found as:
$|E(x,y,z)| \leq \dfrac{1}{2} \times (2) [ |x-1| +|y-1|+|z-2|)^2$
or, $$E \leq \dfrac{1}{2} \times (0.01+0.01+0.08)^2 =0.01$$
