Answer
$0.0024$
Work Step by Step
$f_x=xz-3yz+2 \\ \implies f_x(1,1,2)=(1)(2)-3(1)(2)+2=-2$
Next, $f_y(x,y) =-3z \\ \implies f_{y}(1,1,2) =-6$ and $f_z=x-3y \implies f_z(1,1,2)=1-3(1)=-2 $
$f_{xx}(x,y)=0; f_{yy}(x,y)=0$ and $f_{xy}(x,y) =0$ and $f_{yz}=-3$
The error can be found as:
$|E(x,y,z)| \leq \dfrac{1}{2} \times (3) [ |x-1| +|y-1|+|z-2|)^2$
$\implies E \leq \dfrac{3}{2} \times (0.01+0.01+0.02)^2 =0.0024$
