#### Answer

$$-3.06$$

#### Work Step by Step

Given \begin{aligned}
f(1,0,0) &=-3, & & f_{x}(1,0,0)=-2 \\
f_{y}(1,0,0) &=4, & & f_{z}(1,0,0)=2
\end{aligned}
The linear approximation is given by
\begin{aligned}
f(a+h, b+k,c+r) & \approx f(a, b,c)+hf_{x}(a, b,c) +kf_{y}(a, b,c) +rf_{z}(a,b,c)
\end{aligned}
We find $f (1.02, 0.01, −0.03) $. Here, $h=0.02, k = 0.01,\ r=-0.03$ and $(a,b,c)=(1,0,0)$:
\begin{aligned}
f(1,0,0) & \approx f(1,0,0)+0.002f_{x}(1,0,0)+ 0.01f_{y}(1,0,0)-0.03f_z(1,0,0) \\
& \approx -3-2(0.02)+4(0.01) -2(0.03) \\
& \approx -3.06
\end{aligned}