Question 1207409
**1. Find the normal vector to the plane:**

* We know that the plane contains points A, B, and N. 
* To find the normal vector to the plane, we can use the cross product of two vectors in the plane: AB and AN.

* **Calculate AB:**
  AB = OB - OA = (3i - j - 4k) - (-i + 3j + 5k) = 4i - 4j - 9k

* **Calculate AN:**
  You've already found the position vector of N in part ii). Let's assume you found it to be:
   ON = 1i + 2j + 1k 
  AN = ON - OA = (1i + 2j + 1k) - (-i + 3j + 5k) = 2i - j - 4k

* **Calculate the normal vector (n):**
  n = AB x AN = 
   | i  j  k |
   | 4 -4 -9 |
   | 2 -1 -4 | 
  n = 25i + 10j + 4k

**2. Find the equation of the plane:**

The equation of the plane can be written in the form:

n · (r - A) = 0 

where:
* n is the normal vector to the plane
* r is the position vector of any point on the plane
* A is the position vector of a point on the plane (we can use A or B or N)

Let's use point A:

(25i + 10j + 4k) · (r - (-i + 3j + 5k)) = 0

(25i + 10j + 4k) · (r + i - 3j - 5k) = 0

25x + 10y + 4z + 25 - 30 - 20 = 0

**Therefore, the equation of the plane is:**

25x + 10y + 4z - 25 = 0 

**Note:** 

* Make sure to replace ON with the actual position vector of N that you calculated in part ii) to get the accurate equation of the plane.
* The coefficients of x, y, and z in the equation of the plane represent the components of the normal vector to the plane.