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### **Given Information**:
The equation of the line \( l \) is:
\[
r = (i + pj - 5k) + \lambda(3i - j + qk)
\]
where \( p \) and \( q \) are constants, and \( \lambda \) is a scalar parameter.

The point \( A(-5, 9, -9) \) lies on the line \( l \).

#### **(a) Find the values of \( p \) and \( q \):**

The position vector of a point on the line is given by:
\[
r = (1, p, -5) + \lambda(3, -1, q)
\]
Since point \( A(-5, 9, -9) \) lies on the line, its coordinates should satisfy this equation for some value of \( \lambda \).

We equate the coordinates of \( A \) with those of the line:
- For the \( x \)-coordinate: 
  \[
  -5 = 1 + 3\lambda
  \]
  Solving for \( \lambda \):
  \[
  \lambda = \frac{-5 - 1}{3} = -2
  \]

- For the \( y \)-coordinate:
  \[
  9 = p - \lambda
  \]
  Substituting \( \lambda = -2 \):
  \[
  9 = p - (-2) \implies p = 7
  \]

- For the \( z \)-coordinate:
  \[
  -9 = -5 + q\lambda
  \]
  Substituting \( \lambda = -2 \):
  \[
  -9 = -5 + q(-2) \implies -9 = -5 - 2q \implies -2q = -4 \implies q = 2
  \]

Thus, the values of \( p \) and \( q \) are:
\[
p = 7, \quad q = 2
\]

---

#### **(b) Show that the point \( B(25, -1, 11) \) also lies on \( l \):**

The equation of the line is:
\[
r = (1, 7, -5) + \lambda(3, -1, 2)
\]
We want to verify if \( B(25, -1, 11) \) lies on this line, i.e., there should exist a value of \( \lambda \) such that the coordinates of \( B \) satisfy the parametric equations of the line.

- For the \( x \)-coordinate:
  \[
  25 = 1 + 3\lambda
  \]
  Solving for \( \lambda \):
  \[
  \lambda = \frac{25 - 1}{3} = 8
  \]

- For the \( y \)-coordinate:
  \[
  -1 = 7 - \lambda
  \]
  Substituting \( \lambda = 8 \):
  \[
  -1 = 7 - 8 \implies -1 = -1
  \]
  True.

- For the \( z \)-coordinate:
  \[
  11 = -5 + 2\lambda
  \]
  Substituting \( \lambda = 8 \):
  \[
  11 = -5 + 16 \implies 11 = 11
  \]
  True.

Thus, \( B(25, -1, 11) \) lies on the line \( l \).

---

#### **(c) Find the coordinates of \( C \), where \( OC \) is perpendicular to \( l \):**

We are given that the position vector of \( C \) lies on \( l \), and that \( OC \) is perpendicular to the line \( l \).

Let the direction vector of the line be:
\[
d = (3, -1, 2)
\]

If \( C \) has coordinates \( (x_C, y_C, z_C) \), then:
\[
OC = (x_C, y_C, z_C)
\]
For \( OC \) to be perpendicular to the line, the dot product of \( OC \) and the direction vector of the line must be zero:
\[
OC \cdot d = 0
\]
Substituting the coordinates of \( C \) as \( (1 + 3\lambda, 7 - \lambda, -5 + 2\lambda) \) into the dot product equation:
\[
(1 + 3\lambda) \cdot 3 + (7 - \lambda) \cdot (-1) + (-5 + 2\lambda) \cdot 2 = 0
\]
Expanding the terms:
\[
3(1 + 3\lambda) - (7 - \lambda) + 2(-5 + 2\lambda) = 0
\]
\[
3 + 9\lambda - 7 + \lambda - 10 + 4\lambda = 0
\]
\[
9\lambda + \lambda + 4\lambda + 3 - 7 - 10 = 0
\]
\[
14\lambda - 14 = 0
\]
\[
\lambda = 1
\]

Substitute \( \lambda = 1 \) back into the parametric equations of the line to find the coordinates of \( C \):
\[
x_C = 1 + 3(1) = 4
\]
\[
y_C = 7 - 1 = 6
\]
\[
z_C = -5 + 2(1) = -3
\]
Thus, the coordinates of \( C \) are:
\[
C(4, 6, -3)
\]

---

#### **(d) Find the ratio \( AC : CB \):**

Now, we need to find the lengths of \( AC \) and \( CB \), and then their ratio.

- **Length of \( AC \)**:
  \[
  AC = \sqrt{(4 - (-5))^2 + (6 - 9)^2 + (-3 - (-9))^2}
  \]
  \[
  AC = \sqrt{(4 + 5)^2 + (6 - 9)^2 + (-3 + 9)^2} = \sqrt{9^2 + (-3)^2 + 6^2}
  \]
  \[
  AC = \sqrt{81 + 9 + 36} = \sqrt{126}
  \]
  \[
  AC = \sqrt{126} = 3\sqrt{14}
  \]

- **Length of \( CB \)**:
  \[
  CB = \sqrt{(25 - 4)^2 + (-1 - 6)^2 + (11 - (-3))^2}
  \]
  \[
  CB = \sqrt{(21)^2 + (-7)^2 + (14)^2} = \sqrt{441 + 49 + 196}
  \]
  \[
  CB = \sqrt{686}
  \]

Thus, the ratio \( AC : CB \) is:
\[
\frac{AC}{CB} = \frac{3\sqrt{14}}{\sqrt{686}} = \frac{3\sqrt{14}}{\sqrt{14 \times 49}} = \frac{3\sqrt{14}}{7\sqrt{14}} = \frac{3}{7}
\]

Hence, the ratio \( AC : CB \) is:
\[
AC : CB = 3 : 7
\]

---

Would you like to discuss any step in detail, or go over another problem? Here are 5 related questions you could explore:

1. How can you check if a point lies on a different line?
2. What are the conditions for two vectors to be perpendicular?
3. How do you find the equation of a line passing through two points?
4. What is the formula for the dot product of two vectors in 3D?
5. How do you find the ratio of distances between points on a line?

**Tip**: The dot product is a powerful tool to check orthogonality (perpendicularity) between vectors.

Relative to a fixed origin, O, the line l has the equation r = (i + pj - 5k) + λ(3i - j + qk), where p and q are constants and λ is a scalar parameter. Given that the point A with coordinates (-5, 9, -9) lies on l, (a) find the values of p and q, (b) show that the point B with coordinates (25, -1, 11) also lies on l, (c) find the coordinates of C, where OC is perpendicular to l, (d) find the ratio AC : CB.

Explore how to solve a 3D geometry problem involving a line equation and vector perpendicularity. This problem covers key concepts like parametric equations, dot product, and distance between points, suitable for Grades 11-12 students.

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