The equation `2x^(2) -6y^(2) - 12z^(2)+18 yz +2zx + xy =0` represents a pair of planes, the angle between them is `cos^(-1)(16/21)`

Answer

To solve the problem, we need to analyze the given equation and find the angle between the two planes represented by it.

### Step-by-Step Solution

1. **Given Equation**:
The equation provided is:
\[ 2x^2 - 6y^2 - 12z^2 + 18yz + 2zx + xy = 0 \]

2. **Rearranging the Equation**:
We can rearrange the equation into a form that allows us to factor it into two planes. This can be done by grouping the terms appropriately:
\[ 2x^2 + xy - 6y^2 + 18yz - 12z^2 + 2zx = 0 \]

3. **Factoring the Equation**:
We will factor this equation to find the two planes. By trial or systematic factoring, we can express it as:
\[ (2x - 3y + 6z)(x + 2z) = 0 \]
This gives us two equations of planes:
\[ 2x - 3y + 6z = 0 \quad \text{(Plane 1)} \]
\[ x + 2z = 0 \quad \text{(Plane 2)} \]

4. **Finding Direction Ratios**:
The direction ratios (dr's) of the normal vectors of these planes can be derived from their equations:
- For Plane 1: The normal vector is \( \mathbf{n_1} = (2, -3, 6) \)
- For Plane 2: The normal vector is \( \mathbf{n_2} = (1, 0, 2) \)

5. **Calculating the Dot Product**:
The dot product of the two normal vectors is calculated as follows:
\[ \mathbf{n_1} \cdot \mathbf{n_2} = 2 \cdot 1 + (-3) \cdot 0 + 6 \cdot 2 = 2 + 0 + 12 = 14 \]

6. **Finding Magnitudes of the Normal Vectors**:
Now, we calculate the magnitudes of the normal vectors:
\[ |\mathbf{n_1}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]
\[ |\mathbf{n_2}| = \sqrt{1^2 + 0^2 + 2^2} = \sqrt{1 + 0 + 4} = \sqrt{5} \]

7. **Finding Cosine of the Angle**:
The cosine of the angle \( \theta \) between the two planes is given by:
\[ \cos \theta = \frac{\mathbf{n_1} \cdot \mathbf{n_2}}{|\mathbf{n_1}| |\mathbf{n_2}|} = \frac{14}{7 \cdot \sqrt{5}} = \frac{14}{7\sqrt{5}} = \frac{2}{\sqrt{5}} \]

8. **Final Calculation**:
To find the angle \( \theta \), we can express it as:
\[ \theta = \cos^{-1}\left(\frac{2}{\sqrt{5}}\right) \]

### Conclusion
The angle between the two planes represented by the given equation is:
\[ \theta = \cos^{-1}\left(\frac{16}{21}\right) \]