To calculate the angle formed by the vectors $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}$, we can use the dot product formula:

$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$

Step 1: Calculate the dot product $\vec{a} \cdot \vec{b}$

$\vec{a} \cdot \vec{b} = (2)(6) + (2)(-3) + (-1)(2) = 12 - 6 - 2 = 4$

Step 2: Calculate the magnitudes of $\vec{a}$ and $\vec{b}$

$|\vec{a}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$ $|\vec{b}| = \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$

Step 3: Calculate $\cos \theta$

$\cos \theta = \frac{4}{3 \times 7} = \frac{4}{21}$

Step 4: Find $\theta$ using the inverse cosine

$\theta = \cos^{-1}\left(\frac{4}{21}\right)$

Using a calculator, we get: $\theta \approx 79^\circ$

So, the correct answer is 79°, which corresponds to option A.

Would you like more details or have any questions? Here are some related questions:

1. How do you calculate the magnitude of a vector in three dimensions?
2. What is the significance of the dot product in vector calculations?
3. How does the angle between two vectors affect their dot product?
4. Can the dot product be negative? If so, what does that imply about the angle?
5. How would the calculation change if the vectors were given in two dimensions?

Tip: Remember that the dot product of two vectors is directly related to the cosine of the angle between them, which is why it's useful for finding angles.