(i) Find the co-ordinates of foot of perpendicular drawn from the point (2,3,5) on the plane given by the equation :
2x - 3y + 4z + 10 = 0.
(ii) Find the distance between the point (2,3,-1) and foot of perpendicular drawn from (3,1,-1) to the plane x -y + 3z = 10.

Answer

To solve the given problems step by step, we will break them down into manageable parts.

### Part (i): Finding the coordinates of the foot of the perpendicular from the point (2, 3, 5) to the plane given by the equation \(2x - 3y + 4z + 10 = 0\).

1. **Identify the normal vector of the plane**: The coefficients of \(x\), \(y\), and \(z\) in the plane equation give us the normal vector \(\vec{n} = (2, -3, 4)\).

2. **Write the parametric equations of the line**: The line passing through the point \(P(2, 3, 5)\) and normal to the plane can be expressed as:
\[ \frac{x - 2}{2} = \frac{y - 3}{-3} = \frac{z - 5}{4} = k \]
From this, we can express \(x\), \(y\), and \(z\) in terms of \(k\):
\[ x = 2 + 2k, \quad y = 3 - 3k, \quad z = 5 + 4k \]

3. **Substitute into the plane equation**: Substitute \(x\), \(y\), and \(z\) into the plane equation to find \(k\):
\[ 2(2 + 2k) - 3(3 - 3k) + 4(5 + 4k) + 10 = 0 \]
Simplifying this gives:
\[ 4 + 4k - 9 + 9k + 20 + 16k + 10 = 0 \]
\[ 29k + 25 = 0 \implies k = -\frac{25}{29} \]

4. **Find the coordinates of the foot of the perpendicular**: Substitute \(k\) back into the parametric equations:
\[ x = 2 + 2\left(-\frac{25}{29}\right) = 2 - \frac{50}{29} = \frac{58 - 50}{29} = \frac{8}{29} \]
\[ y = 3 - 3\left(-\frac{25}{29}\right) = 3 + \frac{75}{29} = \frac{87 + 75}{29} = \frac{162}{29} \]
\[ z = 5 + 4\left(-\frac{25}{29}\right) = 5 - \frac{100}{29} = \frac{145 - 100}{29} = \frac{45}{29} \]
Therefore, the coordinates of the foot of the perpendicular are:
\[ \left(\frac{8}{29}, \frac{162}{29}, \frac{45}{29}\right) \]

### Part (ii): Finding the distance between the point (2, 3, -1) and the foot of the perpendicular drawn from (3, 1, -1) to the plane \(x - y + 3z = 10\).

1. **Identify the normal vector of the plane**: The normal vector \(\vec{n} = (1, -1, 3)\).

2. **Write the parametric equations of the line**: The line passing through the point \(Q(3, 1, -1)\) and normal to the plane can be expressed as:
\[ \frac{x - 3}{1} = \frac{y - 1}{-1} = \frac{z + 1}{3} = k \]
From this, we can express \(x\), \(y\), and \(z\) in terms of \(k\):
\[ x = 3 + k, \quad y = 1 - k, \quad z = -1 + 3k \]

3. **Substitute into the plane equation**: Substitute \(x\), \(y\), and \(z\) into the plane equation to find \(k\):
\[ (3 + k) - (1 - k) + 3(-1 + 3k) = 10 \]
Simplifying this gives:
\[ 3 + k - 1 + k - 3 + 9k = 10 \]
\[ 11k - 1 = 10 \implies 11k = 11 \implies k = 1 \]

4. **Find the coordinates of the foot of the perpendicular**: Substitute \(k\) back into the parametric equations:
\[ x = 3 + 1 = 4, \quad y = 1 - 1 = 0, \quad z = -1 + 3(1) = 2 \]
Therefore, the coordinates of the foot of the perpendicular are:
\[ (4, 0, 2) \]

5. **Calculate the distance**: Use the distance formula between the points \( (2, 3, -1) \) and \( (4, 0, 2) \):
\[ d = \sqrt{(4 - 2)^2 + (0 - 3)^2 + (2 - (-1))^2} \]
\[ = \sqrt{2^2 + (-3)^2 + 3^2} = \sqrt{4 + 9 + 9} = \sqrt{22} \]

### Final Answers:
1. The coordinates of the foot of the perpendicular from the point (2, 3, 5) to the plane \(2x - 3y + 4z + 10 = 0\) are \(\left(\frac{8}{29}, \frac{162}{29}, \frac{45}{29}\right)\).
2. The distance between the point (2, 3, -1) and the foot of the perpendicular from (3, 1, -1) to the plane \(x - y + 3z = 10\) is \(\sqrt{22}\).