Answer
$0.00135$
Work Step by Step
$f_x=y-3z \\ \implies f_x(1,1,0)=1-3(0)=1$
Next, $f_y =x+2z \\ \implies f_{y}(1,1,0) =x+2z=1+2(0)=1$ and $f_z=2y-3x \implies f_z(1,1,0)=2y-3x=2(1)-3(1)=-1$
$f_{xx}=0; f_{yy}=0$ and $f_{zz} =0$ and $f_{xy}=1$ and $f_{xz}=-3$ and $f_{yz}=2$
The error can be found as:
$|E(x,y,z)| \leq \dfrac{1}{2} \times (3) [ |x-1| +|y-1|+|z-0|)^2$
$\implies E \leq \dfrac{3}{2} \times (0.01+0.01+0.01)^2 =0.00135$
