(i) Find the distance of the point P (6,5,9) from the plane determined by the points A (3,-1,2) B (5,2,4) and C (-1,-1,6).
(ii) Find the distance between the point (7,2,4) and the plane determined by the points.
A (2,5,-3) B (-2,-3,5) and C (5,3,-3,) .

Answer

To solve the given problems step by step, we will follow the method of finding the distance from a point to a plane defined by three points in three-dimensional space.

### Part (i): Find the distance of the point P (6,5,9) from the plane determined by the points A (3,-1,2), B (5,2,4), and C (-1,-1,6).

**Step 1: Find the vectors AB and AC.**
- \( \vec{AB} = B - A = (5 - 3, 2 - (-1), 4 - 2) = (2, 3, 2) \)
- \( \vec{AC} = C - A = (-1 - 3, -1 - (-1), 6 - 2) = (-4, 0, 4) \)

**Step 2: Find the normal vector to the plane using the cross product of AB and AC.**
- \( \vec{n} = \vec{AB} \times \vec{AC} \)
- The determinant for the cross product is given by:
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 2 \\ -4 & 0 & 4 \end{vmatrix} \]
Calculating this determinant:
\[ \vec{n} = \hat{i}(3 \cdot 4 - 0 \cdot 2) - \hat{j}(2 \cdot 4 - 2 \cdot (-4)) + \hat{k}(2 \cdot 0 - 3 \cdot (-4)) \]
\[ = \hat{i}(12) - \hat{j}(8 + 8) + \hat{k}(0 + 12) \]
\[ = 12\hat{i} - 16\hat{j} + 12\hat{k} \]
Thus, \( \vec{n} = (12, -16, 12) \).

**Step 3: Find the equation of the plane.**
Using point A (3, -1, 2) and the normal vector \( \vec{n} \):
\[ 12(x - 3) - 16(y + 1) + 12(z - 2) = 0 \]
Expanding this gives:
\[ 12x - 36 - 16y - 16 + 12z - 24 = 0 \]
\[ 12x - 16y + 12z - 76 = 0 \]

**Step 4: Find the distance from point P (6, 5, 9) to the plane.**
The distance \( d \) from point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
Here, \( A = 12, B = -16, C = 12, D = -76 \), and \( P(6, 5, 9) \):
\[ d = \frac{|12 \cdot 6 - 16 \cdot 5 + 12 \cdot 9 - 76|}{\sqrt{12^2 + (-16)^2 + 12^2}} \]
Calculating the numerator:
\[ = |72 - 80 + 108 - 76| = |24| = 24 \]
Calculating the denominator:
\[ = \sqrt{144 + 256 + 144} = \sqrt{544} = 4\sqrt{34} \]
Thus, the distance is:
\[ d = \frac{24}{4\sqrt{34}} = \frac{6}{\sqrt{34}} \]

### Part (ii): Find the distance between the point (7,2,4) and the plane determined by the points A (2,5,-3), B (-2,-3,5), and C (5,3,-3).

**Step 1: Find the vectors AB and AC.**
- \( \vec{AB} = B - A = (-2 - 2, -3 - 5, 5 - (-3)) = (-4, -8, 8) \)
- \( \vec{AC} = C - A = (5 - 2, 3 - 5, -3 - (-3)) = (3, -2, 0) \)

**Step 2: Find the normal vector to the plane using the cross product of AB and AC.**
- \( \vec{n} = \vec{AB} \times \vec{AC} \)
- The determinant for the cross product is given by:
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & -8 & 8 \\ 3 & -2 & 0 \end{vmatrix} \]
Calculating this determinant:
\[ \vec{n} = \hat{i}((-8) \cdot 0 - 8 \cdot (-2)) - \hat{j}((-4) \cdot 0 - 8 \cdot 3) + \hat{k}((-4) \cdot (-2) - (-8) \cdot 3) \]
\[ = \hat{i}(16) - \hat{j}(-24) + \hat{k}(8 + 24) \]
\[ = 16\hat{i} + 24\hat{j} + 32\hat{k} \]
Thus, \( \vec{n} = (16, 24, 32) \).

**Step 3: Find the equation of the plane.**
Using point A (2, 5, -3) and the normal vector \( \vec{n} \):
\[ 16(x - 2) + 24(y - 5) + 32(z + 3) = 0 \]
Expanding this gives:
\[ 16x - 32 + 24y - 120 + 32z + 96 = 0 \]
\[ 16x + 24y + 32z - 56 = 0 \]

**Step 4: Find the distance from point (7, 2, 4) to the plane.**
Using the formula for distance:
\[ d = \frac{|16 \cdot 7 + 24 \cdot 2 + 32 \cdot 4 - 56|}{\sqrt{16^2 + 24^2 + 32^2}} \]
Calculating the numerator:
\[ = |112 + 48 + 128 - 56| = |232| = 232 \]
Calculating the denominator:
\[ = \sqrt{256 + 576 + 1024} = \sqrt{1856} = 16\sqrt{7} \]
Thus, the distance is:
\[ d = \frac{232}{16\sqrt{7}} = \frac{29}{2\sqrt{7}} \]

### Final Answers:
1. The distance from point P (6,5,9) to the plane is \( \frac{6}{\sqrt{34}} \).
2. The distance from point (7,2,4) to the plane is \( \frac{29}{2\sqrt{7}} \).