Discussion
The tangent plane to \( x^{2} + y^{2} + z^{2} = 9 \) at \( (1, 2, 2) \) has equation
Answer: A
The gradient \( 2\mathbf{i} + 4\mathbf{j} + 4\mathbf{k} \) is normal there, and dividing by two gives the simpler normal \( (1, 2, 2) \). The plane through the point with that normal is
\[ 1(x-1) + 2(y-2) + 2(z-2) \;=\; 0 \quad \Longrightarrow \quad x + 2y + 2z \;=\; 9. \]
Substituting the point confirms \( 1 + 4 + 4 = 9 \)
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