If `overline(P_1P_2)` is perpendicular to `overline(P_2P_3)`, then the value of k is, where `P_1(k, 1, -1), P_2(2k, 0, 2) and P_3(2+2k, k, 1)` is …..

Answer

To solve the problem, we need to find the value of \( k \) such that the vector \( \overline{P_1P_2} \) is perpendicular to the vector \( \overline{P_2P_3} \).

### Step 1: Define the Points
We have the points:
- \( P_1(k, 1, -1) \)
- \( P_2(2k, 0, 2) \)
- \( P_3(2 + 2k, k, 1) \)

### Step 2: Find the Vector \( \overline{P_1P_2} \)
The vector \( \overline{P_1P_2} \) can be calculated as:
\[ \overline{P_1P_2} = P_2 - P_1 = (2k - k, 0 - 1, 2 - (-1)) \]
This simplifies to:
\[ \overline{P_1P_2} = (k, -1, 3) \]

### Step 3: Find the Vector \( \overline{P_2P_3} \)
Next, we calculate the vector \( \overline{P_2P_3} \):
\[ \overline{P_2P_3} = P_3 - P_2 = ((2 + 2k) - 2k, k - 0, 1 - 2) \]
This simplifies to:
\[ \overline{P_2P_3} = (2, k, -1) \]

### Step 4: Use the Perpendicular Condition
For the vectors to be perpendicular, their dot product must equal zero:
\[ \overline{P_1P_2} \cdot \overline{P_2P_3} = 0 \]
Calculating the dot product:
\[ (k, -1, 3) \cdot (2, k, -1) = k \cdot 2 + (-1) \cdot k + 3 \cdot (-1) \]
This expands to:
\[ 2k - k - 3 = 0 \]
Simplifying this gives:
\[ k - 3 = 0 \]

### Step 5: Solve for \( k \)
From the equation \( k - 3 = 0 \), we find:
\[ k = 3 \]

### Final Answer
Thus, the value of \( k \) is \( 3 \).

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