Using vectors, find the value of k, such that the points (k,-10, 3), (1,-1, 3) and (3, 5, 3) are collinear.

Explanation

To find the value of \( k \) such that the points \( A(k, -10, 3) \), \( B(1, -1, 3) \), and \( C(3, 5, 3) \) are collinear, we can use the concept of vectors. Here’s a step-by-step solution:

### Step 1: Define the Points
Let:
- \( A = (k, -10, 3) \)
- \( B = (1, -1, 3) \)
- \( C = (3, 5, 3) \)

### Step 2: Find the Vectors \( \overrightarrow{AB} \) and \( \overrightarrow{BC} \)
The vector \( \overrightarrow{AB} \) can be calculated as:
\[ \overrightarrow{AB} = B - A = (1 - k, -1 - (-10), 3 - 3) = (1 - k, 9, 0) \]

The vector \( \overrightarrow{BC} \) can be calculated as:
\[ \overrightarrow{BC} = C - B = (3 - 1, 5 - (-1), 3 - 3) = (2, 6, 0) \]

### Step 3: Set Up the Collinearity Condition
For the points to be collinear, the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{BC} \) must be parallel. This means that:
\[ \overrightarrow{AB} = \lambda \overrightarrow{BC} \]
for some scalar \( \lambda \).

### Step 4: Write the Equation
From the vectors we found:
\[ (1 - k, 9, 0) = \lambda (2, 6, 0) \]

This gives us the following equations:
1. \( 1 - k = 2\lambda \)
2. \( 9 = 6\lambda \)

### Step 5: Solve for \( \lambda \)
From the second equation:
\[ 9 = 6\lambda \implies \lambda = \frac{9}{6} = \frac{3}{2} \]

### Step 6: Substitute \( \lambda \) Back
Substituting \( \lambda = \frac{3}{2} \) into the first equation:
\[ 1 - k = 2 \left(\frac{3}{2}\right) \]
\[ 1 - k = 3 \]

### Step 7: Solve for \( k \)
Rearranging gives:
\[ -k = 3 - 1 \implies -k = 2 \implies k = -2 \]

### Conclusion
The value of \( k \) such that the points \( A(k, -10, 3) \), \( B(1, -1, 3) \), and \( C(3, 5, 3) \) are collinear is:
\[ \boxed{-2} \]