Given surfaces are

z = x² + y² – 3

\(\Rightarrow\) x² + y² – z = 3

And x² + y² + z² = 9

let \(\phi_1\) = x² + y² – z

And \(\phi_2\) = x² + y² + z²

Given point (2, –1, 2).

\(\vec \nabla \phi_1 = \vec \nabla(\mathrm x^2+y^2-z)\)

\(=\hat i\frac{\partial}{\partial \mathrm x}(\mathrm x^2+y^2-z)\) \(+\hat j\frac{\partial}{\partial \mathrm x}(\mathrm x^2+y^2-z)\) \(+\hat k \frac{\partial }{\partial z}(\mathrm x^2+y^2-z)\)

\(=2\mathrm x \hat i + 2y\hat j-\hat k\)

\((\vec \nabla\phi_1)_{(2, -1, 2) }= 4 \hat i -2\hat j-\hat k\)

\(\vec \nabla \phi_2=\vec \nabla(\mathrm x^2+y^2+z^2)\)

\(=\hat i\frac{\partial}{\partial \mathrm x}(\mathrm x^2+y^2+z^2)\) \(+\hat j \frac{\partial}{\partial y}(\mathrm x^2+y^2+z^2)\) \(+\hat k\frac{\partial}{\partial z}(\mathrm x^2+y^2+z^2)\)

\(=2\mathrm x \hat i+2y\hat j + 2z\hat k\)

\((\vec \nabla \phi_2)_{(2, -1, 2)}=4\hat i-2\hat j+4\hat k\)

let angle between surfaces \(\phi_1\) and \(\phi_2\) is θ.

Then \(\vec \nabla\phi_1.\vec \nabla \phi_2\) \(=|\vec \nabla \phi_1||\vec \nabla \phi_2|\cos \theta\)

\(\Rightarrow \cos\theta = \frac{\vec \Delta \phi_1.\vec \nabla\phi_2}{|\vec \nabla \phi_1||\vec \nabla.\phi_2|}\)

\(=\frac{(4\hat i-2\hat j-\hat k)(4\hat i-2\hat j+4\hat k)}{|4\hat i-2\hat j-\hat k||4\hat i-2\hat j+4\hat k|}\)

\(=\frac{16+4-4}{\sqrt{16+4+1}\sqrt{16+4+16}}\)    \(\Big(\because \hat i.\hat i=\hat j.\hat j=\hat k.\hat k=1\) & \(\hat i.\hat j=\hat i.\hat k=\hat j.\hat k=0\Big)\)

\(=\frac{16}{\sqrt{21}\sqrt{36}}=\frac{16}{6\sqrt{21}}=\frac{8}{3\sqrt{21}}\)

\(\Rightarrow \theta =\cos^{-1}\left(\frac{8}{3\sqrt{21}}\right)\)

Hence, angle between given surfaces is \(\cos^{-1}\frac{8}{3\sqrt{21}}\).