If the points `A(2,1,1,),B(0,-1,4)andC(k,3,-2)` are collinear, then k= . . . . . . . .

Options

A. 0
B. 1
C. 4
D. `-4`

Answer

To determine the value of \( k \) for which the points \( A(2, 1, 1) \), \( B(0, -1, 4) \), and \( C(k, 3, -2) \) are collinear, we will follow these steps:

### Step 1: Find the direction vectors
First, we need to find the direction vectors \( \overrightarrow{AB} \) and \( \overrightarrow{BC} \).

The direction vector \( \overrightarrow{AB} \) from point \( A \) to point \( B \) is given by:
\[ \overrightarrow{AB} = B - A = (0 - 2, -1 - 1, 4 - 1) = (-2, -2, 3) \]

The direction vector \( \overrightarrow{BC} \) from point \( B \) to point \( C \) is given by:
\[ \overrightarrow{BC} = C - B = (k - 0, 3 - (-1), -2 - 4) = (k, 4, -6) \]

### Step 2: Set up the condition for collinearity
For the points to be collinear, the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{BC} \) must be parallel. This means that there exists a scalar \( \lambda \) such that:
\[ \overrightarrow{AB} = \lambda \overrightarrow{BC} \]

This gives us the following equations:
\[ -2 = \lambda k \quad (1) \]
\[ -2 = \lambda \cdot 4 \quad (2) \]
\[ 3 = \lambda \cdot (-6) \quad (3) \]

### Step 3: Solve for \( \lambda \)
From equation (2):
\[ \lambda = \frac{-2}{4} = -\frac{1}{2} \]

### Step 4: Substitute \( \lambda \) into the other equations
Now, substitute \( \lambda = -\frac{1}{2} \) into equation (1):
\[ -2 = -\frac{1}{2} k \]
Multiplying both sides by -2 gives:
\[ 4 = k \quad (4) \]

### Step 5: Verify with the third equation
Now, let's verify with equation (3):
\[ 3 = -\frac{1}{2} \cdot (-6) \]
This simplifies to:
\[ 3 = 3 \quad (True) \]

Thus, all conditions are satisfied.

### Conclusion
The value of \( k \) for which the points \( A \), \( B \), and \( C \) are collinear is:
\[ \boxed{4} \]