Angle between the planes:
(i) `vec(r). (hati - 2 hatj - hatk) = 1 and vec(r). (3 hati - 6 hatj + 2 hatk)` = 0
(ii) `vec(r). (2 hati + 2 hatj - 3 hatk ) = 5 and vec(r) . ( 3 hati - 3 hatj + 5 hatk ) = 3 `

Answer

To find the angle between the given planes, we will follow these steps:

### Step 1: Identify the normal vectors of the planes

For the first set of planes:
1. The equation of the first plane is given as:
\[ \vec{r} \cdot (\hat{i} - 2\hat{j} - \hat{k}) = 1 \]
The normal vector \( \vec{n_1} \) is:
\[ \vec{n_1} = \hat{i} - 2\hat{j} - \hat{k} = (1, -2, -1) \]

2. The equation of the second plane is given as:
\[ \vec{r} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 0 \]
The normal vector \( \vec{n_2} \) is:
\[ \vec{n_2} = 3\hat{i} - 6\hat{j} + 2\hat{k} = (3, -6, 2) \]

### Step 2: Use the formula for the angle between two planes

The angle \( \theta \) between the two planes can be calculated using the formula:
\[ \cos \theta = \frac{\vec{n_1} \cdot \vec{n_2}}{|\vec{n_1}| |\vec{n_2}|} \]

### Step 3: Calculate the dot product \( \vec{n_1} \cdot \vec{n_2} \)

Calculating the dot product:
\[ \vec{n_1} \cdot \vec{n_2} = (1)(3) + (-2)(-6) + (-1)(2) = 3 + 12 - 2 = 13 \]

### Step 4: Calculate the magnitudes of \( \vec{n_1} \) and \( \vec{n_2} \)

Calculating the magnitude of \( \vec{n_1} \):
\[ |\vec{n_1}| = \sqrt{1^2 + (-2)^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \]

Calculating the magnitude of \( \vec{n_2} \):
\[ |\vec{n_2}| = \sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7 \]

### Step 5: Substitute values into the cosine formula

Now substituting back into the cosine formula:
\[ \cos \theta = \frac{13}{\sqrt{6} \cdot 7} = \frac{13}{7\sqrt{6}} \]

### Step 6: Find the angle \( \theta \)

Finally, we find \( \theta \):
\[ \theta = \cos^{-1}\left(\frac{13}{7\sqrt{6}}\right) \]

### Step 7: Repeat for the second set of planes

For the second set of planes:
1. The normal vector \( \vec{n_1} = (2, 2, -3) \)
2. The normal vector \( \vec{n_2} = (3, -3, 5) \)

Calculating the dot product:
\[ \vec{n_1} \cdot \vec{n_2} = (2)(3) + (2)(-3) + (-3)(5) = 6 - 6 - 15 = -15 \]

Calculating the magnitudes:
\[ |\vec{n_1}| = \sqrt{2^2 + 2^2 + (-3)^2} = \sqrt{4 + 4 + 9} = \sqrt{17} \]
\[ |\vec{n_2}| = \sqrt{3^2 + (-3)^2 + 5^2} = \sqrt{9 + 9 + 25} = \sqrt{43} \]

Substituting into the cosine formula:
\[ \cos \theta = \frac{-15}{\sqrt{17} \cdot \sqrt{43}} \]

Finding the angle \( \theta \):
\[ \theta = \cos^{-1}\left(\frac{-15}{\sqrt{731}}\right) \]