To solve the problem, we will break it down into two parts as given in the question.
### Part (a)
#### (i) Finding the foot of the perpendicular from the point (2, -1, 5) on the line:
The line is given in symmetric form:
\[
\frac{x - 11}{10} = \frac{y + 2}{-5} = \frac{z + 8}{11}
\]
1. **Identify the point on the line and the direction ratios**:
- The line passes through the point \( (11, -2, -8) \) and has direction ratios \( (10, -5, 11) \).
- Let \( D \) be the foot of the perpendicular from point \( P(2, -1, 5) \) to the line.
2. **Parameterize the line**:
- Let \( \lambda \) be the parameter. The coordinates of point \( D \) on the line can be expressed as:
\[
D(\lambda) = (10\lambda + 11, -5\lambda - 2, 11\lambda - 8)
\]
3. **Find the vector \( PD \)**:
- The vector \( PD \) is given by:
\[
PD = D(\lambda) - P = (10\lambda + 11 - 2, -5\lambda - 2 + 1, 11\lambda - 8 - 5)
\]
- Simplifying this gives:
\[
PD = (10\lambda + 9, -5\lambda - 1, 11\lambda - 13)
\]
4. **Direction vector of the line**:
- The direction vector \( \mathbf{b} \) of the line is \( (10, -5, 11) \).
5. **Set up the dot product equation**:
- Since \( PD \) is perpendicular to \( \mathbf{b} \), we have:
\[
PD \cdot \mathbf{b} = 0
\]
- This gives:
\[
(10\lambda + 9) \cdot 10 + (-5\lambda - 1) \cdot (-5) + (11\lambda - 13) \cdot 11 = 0
\]
6. **Solve the equation**:
- Expanding and simplifying:
\[
100\lambda + 90 + 25\lambda + 5 + 121\lambda - 143 = 0
\]
- Combine like terms:
\[
246\lambda - 48 = 0 \implies \lambda = \frac{48}{246} = \frac{8}{41}
\]
7. **Substitute \( \lambda \) back to find coordinates of \( D \)**:
- Substitute \( \lambda \) into \( D(\lambda) \):
\[
D = \left(10 \cdot \frac{8}{41} + 11, -5 \cdot \frac{8}{41} - 2, 11 \cdot \frac{8}{41} - 8\right)
\]
- Calculate each coordinate:
\[
D = \left(\frac{80}{41} + \frac{451}{41}, -\frac{40}{41} - \frac{82}{41}, \frac{88}{41} - \frac{328}{41}\right)
\]
\[
D = \left(\frac{531}{41}, -\frac{122}{41}, -\frac{240}{41}\right)
\]
#### (ii) Finding the foot of the perpendicular from the point (0, 2, 3) on the line:
The line is given as:
\[
\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}
\]
1. **Identify the point on the line and the direction ratios**:
- The line passes through the point \( (-3, 1, -4) \) and has direction ratios \( (5, 2, 3) \).
- Let \( D' \) be the foot of the perpendicular from point \( P'(0, 2, 3) \) to the line.
2. **Parameterize the line**:
- Let \( \mu \) be the parameter. The coordinates of point \( D' \) on the line can be expressed as:
\[
D'(\mu) = (5\mu - 3, 2\mu + 1, 3\mu - 4)
\]
3. **Find the vector \( PD' \)**:
- The vector \( PD' \) is given by:
\[
PD' = D'(\mu) - P' = (5\mu - 3 - 0, 2\mu + 1 - 2, 3\mu - 4 - 3)
\]
- Simplifying gives:
\[
PD' = (5\mu - 3, 2\mu - 1, 3\mu - 7)
\]
4. **Direction vector of the line**:
- The direction vector \( \mathbf{b'} \) of the line is \( (5, 2, 3) \).
5. **Set up the dot product equation**:
- Since \( PD' \) is perpendicular to \( \mathbf{b'} \), we have:
\[
PD' \cdot \mathbf{b'} = 0
\]
- This gives:
\[
(5\mu - 3) \cdot 5 + (2\mu - 1) \cdot 2 + (3\mu - 7) \cdot 3 = 0
\]
6. **Solve the equation**:
- Expanding and simplifying:
\[
25\mu - 15 + 4\mu - 2 + 9\mu - 21 = 0
\]
- Combine like terms:
\[
38\mu - 38 = 0 \implies \mu = 1
\]
7. **Substitute \( \mu \) back to find coordinates of \( D' \)**:
- Substitute \( \mu \) into \( D'(\mu) \):
\[
D' = (5 \cdot 1 - 3, 2 \cdot 1 + 1, 3 \cdot 1 - 4)
\]
- Calculate each coordinate:
\[
D' = (2, 3, -1)
\]
### Part (b)
#### Finding the length of the perpendicular in part (ii):
1. **Calculate the distance between points \( P' \) and \( D' \)**:
- The distance formula in 3D is given by:
\[
\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}
\]
- Here, \( P' = (0, 2, 3) \) and \( D' = (2, 3, -1) \):
\[
\text{Distance} = \sqrt{(2 - 0)^2 + (3 - 2)^2 + (-1 - 3)^2}
\]
\[
= \sqrt{4 + 1 + 16} = \sqrt{21}
\]
### Final Answers:
- (i) The foot of the perpendicular from point \( (2, -1, 5) \) is \( \left( \frac{531}{41}, -\frac{122}{41}, -\frac{240}{41} \right) \).
- (ii) The foot of the perpendicular from point \( (0, 2, 3) \) is \( (2, 3, -1) \).
- (b) The length of the perpendicular in part (ii) is \( \sqrt{21} \).
