Answer
$L(x,y,z)=\dfrac{3}{7}x+\dfrac{2}{7}y+\dfrac{6}{7}z$ and $6.9914$
Work Step by Step
We need to find the partial derivatives w.r.t. $x$, $y$, and $z$ as follows:
$f_x=\dfrac{1}{2}(x^2+y^2+z^2)^{-(1/2)} \cdot (2x)=\dfrac{x}{\sqrt{x^2+y^2+z^2}};f_y=\dfrac{y}{\sqrt{x^2+y^2+z^2}}$
and $f_z=\dfrac{z}{\sqrt{x^2+y^2+z^2}}$
The partial derivatives at the point $(3,2,6)$ are:
$w_0=f(3,2,6)=\sqrt{3^2+2^2+6^2}=7\\
f_x(3,2,6)=\dfrac{3}{7}\\f_y(3,2,6)=\dfrac{2}{7}\\f_z(3,2,6)=\dfrac{6}{7}$
Linear approximation to $f$ can be defined as:
$L(x,y,z)=\dfrac{3}{7}(x-3)+\dfrac{2}{7}(y-2)+\dfrac{6}{7}(z-6)+7=\dfrac{3}{7}x+\dfrac{2}{7}y+\dfrac{6}{7}z$ +7
and
$L(3.02,1.97,5.99)=\dfrac{3}{7}(3.02-3)+\dfrac{2}{7}(1.97-2)+\dfrac{6}{7}(5.99-6) \approx 6.9914$
