Let $A = x^2 + y^2 + z^2 = 9 \\ B = x^2 + y^2 - z = 3$

Taking gradient of the given surfaces

$\bar{V}A = \frac{\partial A}{\partial x}\hat{i} + \frac{\partial A}{\partial y}\hat{j} + \frac{\partial A}{\partial z}\hat{k} = 2x\hat{i} + 2y\hat{j} + 2z\hat{k} = \bar{C}$

Similarly,

$\bar{V}B = 2x\hat{i} + 2y\hat{j} - \hat{k} = \bar{D}$

The vectors at point (2, -1, 2) is

$\therefore \bar{C} = 4\hat{i} + 2(-1)\hat{j} + 2(2)\hat{k}$

Now taking dot product of vectors C and D we get

$\bar{C} \cdot \bar{D} = |C| |D| cos\theta \\ \therefore (4\hat{i} - 2\hat{j} + 4\hat{k})(4\hat{i} - 2\hat{j} - \hat{k}) = \sqrt{4^2 + 2^2 + 4^2}\sqrt{4^2 + 2^2 + 1^2}cos\theta \\ \therefore 16 + 4 - 4 = \sqrt{36}\sqrt{21}cos\theta \\ 16 = 6\sqrt{21}cos\theta \\ \therefore cos\theta = \frac{8}{3\sqrt{21}} \\ \therefore \theta = cos^{-1}\bigg(\frac{8}{3\sqrt{21}}\bigg) \\ \therefore \theta = 54.41^{\circ}$

The angle between two surfaces $x^2 + y^2 + z^2 = 9$ and $x^2 + y^2 - z = 3$ at point (2, - 1, 2) is $cos^{-1}\bigg(\frac{8}{3\sqrt{21}}\bigg)$