Given, \(\vec a = 3\hat i + 2\hat j + 2\hat k\)
and \(\vec b = \hat i + 2\hat j - 2\hat k\)
Now, \(\vec a + \vec b = (3\hat i + 2\hat j + 2\hat k) + (\hat i + 2\hat j - 2\hat k)= 4 \hat i + 4\hat j + 0\hat k\)
\(\vec a - \vec b = (3\hat i + 2\hat j + 2\hat k) - (\hat i + 2\hat j - 2\hat k) = 2\hat i + 0\hat j + 4\hat k\)
We know that the unit vector perpendicular to both \(\vec a + \vec b\) and \(\vec a -\vec b\) is given by
\(\hat n = \frac{(\vec a + \vec b)\times (\vec a - \vec b)}{|(\vec a + \vec b)\times (\vec a - \vec b)|}\)
Here, \((\vec a + \vec b) \times (\vec a - \vec b) = \begin{vmatrix}\hat i &\hat j &\hat k\\4&4&0\\2&0&4\end{vmatrix}\)
\(= \hat i (16 - 0) - \hat j (16 - 0) + \hat k (0-8)\)
\(= 16 \hat i - 16\hat j - 8\hat k\)
Now, \(|(\vec a + \vec b) \times (\vec a - \vec b)| = \sqrt{(16)^2 + (16)^2 + (8)^2}\)
\(= \sqrt{256 + 256 + 64}\)
\(= \sqrt{576}\)
\(= 24\)
Hence, the required unit vector \(\hat n = \frac{(\vec a + \vec b)\times (\vec a - \vec b)}{|(\vec a + \vec b)\times (\vec a - \vec b)|}\)
\(=\frac{16\hat i - 16\hat j - 8\hat k}{24}\)
\(= \frac 23 \hat i - \frac 23 \hat j - \frac 13 \hat k\)
