Answer
$\dfrac{3}{7}x+\dfrac{2}{7}y+\dfrac{6}{7}z$
and
$L(3.02,1.97,5.99) \approx 6.9914$
Work Step by Step
The partial derivatives of the given function with respect to $x$, $y$, and $z$ are given as follows:
$f_x=(2x) \times \dfrac{1}{2}(x^2+y^2+z^2)^{-(1/2)} =\dfrac{x}{\sqrt{x^2+y^2+z^2}}$ and $f_y=\dfrac{y}{\sqrt{x^2+y^2+z^2}}
;f_z=\dfrac{z}{\sqrt{x^2+y^2+z^2}}$
and $f_x(3,2,6)=\dfrac{3}{7}\\f_y(3,2,6)=\dfrac{2}{7}\\f_z(3,2,6)=\dfrac{6}{7}$
The partial derivatives at the point $(3,2,6)$ are:
$W_0=f(3,2,6)=\sqrt{(3)^2+(2)^2+(6)^2}=\sqrt{9+4+36}=7$
Linear approximation to $f$ can be defined as:
$L(x,y,z)=\dfrac{3}{7} \times (x-3)+\dfrac{2}{7} \times (y-2)+\dfrac{6}{7} \times (z-6)+7=\dfrac{3}{7}x+\dfrac{2}{7}y+\dfrac{6}{7}z$
and
$L(3.02,1.97,5.99)=\dfrac{3}{7} \times (3.02)+\dfrac{2}{7} \times (1.97)+\dfrac{6}{7} \times (5.99) \approx 6.9914$
