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Consider the function

$\displaystyle f(x,y,z)=x^2+y^2+z^2-9$

which has partial derivatives

$\displaystyle f_x(x,y,z)=2x,\ f_y(x,y,z)=2y,\ f_z(x,y,z)=2z$

So if we consider its first Taylor polynomial about the point $\displaystyle (1,2,2)$ :

$\displaystyle P(x,y,z)=f(1,2,2)+f_x(1,2,2)(x-1)+f_y(1,2,2)(y-2)+f_z(1,2,2)(z-2) $

$\displaystyle = 2(x-1)+4(y-2)+4(z-2)$

Now notice that your sphere is the solution space of $\displaystyle f(x,y,z)=0$; and the tangent plane at the point $\displaystyle (1,2,2)$ will be the solution space of $\displaystyle P(x,y,z)=0$, i.e.

$\displaystyle 2(x-1)+4(y-2)+4(z-2)=0$

or

$\displaystyle x+2y+2z=9.$