A line passes through the point `A(2, 3, 5)` and is parallel to the vector `2hati-hatj+hatk`. If P is a point on this line such that `AP=2sqrt6`, then the coordinates of point P can be

Options

A. `(4, 2, 6)`
B. `(6, 1, 7)`
C. `(-2, 5, -3)`
D. `(2, 3, 5)`

Answer

To find the coordinates of point P that lies on the line passing through point A(2, 3, 5) and is parallel to the vector \( \mathbf{v} = 2\hat{i} - \hat{j} + \hat{k} \), we can follow these steps:

### Step 1: Write the parametric equations of the line
The line can be represented in parametric form. Since it passes through point A(2, 3, 5) and is parallel to the vector \( \mathbf{v} = 2\hat{i} - \hat{j} + \hat{k} \), we can express the coordinates of any point P on the line as:
\[ P(x, y, z) = (2, 3, 5) + \lambda(2, -1, 1) \]
This gives us the parametric equations:
\[ x = 2 + 2\lambda \]
\[ y = 3 - \lambda \]
\[ z = 5 + \lambda \]

### Step 2: Find the distance AP
The distance \( AP \) is given as \( 2\sqrt{6} \). The vector \( \overrightarrow{AP} \) can be expressed as:
\[ \overrightarrow{AP} = P - A = (x - 2, y - 3, z - 5) \]
Substituting the parametric equations into this, we have:
\[ \overrightarrow{AP} = (2 + 2\lambda - 2, 3 - \lambda - 3, 5 + \lambda - 5) = (2\lambda, -\lambda, \lambda) \]

### Step 3: Calculate the magnitude of AP
The magnitude of vector \( \overrightarrow{AP} \) is:
\[ |\overrightarrow{AP}| = \sqrt{(2\lambda)^2 + (-\lambda)^2 + (\lambda)^2} \]
Calculating this gives:
\[ |\overrightarrow{AP}| = \sqrt{4\lambda^2 + \lambda^2 + \lambda^2} = \sqrt{6\lambda^2} = \sqrt{6}|\lambda| \]

### Step 4: Set the magnitude equal to the given distance
We set the magnitude equal to \( 2\sqrt{6} \):
\[ \sqrt{6}|\lambda| = 2\sqrt{6} \]
Dividing both sides by \( \sqrt{6} \):
\[ |\lambda| = 2 \]

### Step 5: Solve for \( \lambda \)
This gives us two possible values for \( \lambda \):
\[ \lambda = 2 \quad \text{or} \quad \lambda = -2 \]

### Step 6: Find the coordinates of point P for each \( \lambda \)
1. For \( \lambda = 2 \):
\[ x = 2 + 2(2) = 6 \]
\[ y = 3 - 2 = 1 \]
\[ z = 5 + 2 = 7 \]
So, one possible point P is \( (6, 1, 7) \).

2. For \( \lambda = -2 \):
\[ x = 2 + 2(-2) = -2 \]
\[ y = 3 - (-2) = 5 \]
\[ z = 5 + (-2) = 3 \]
So, another possible point P is \( (-2, 5, 3) \).

### Final Answer
The coordinates of point P can be \( (6, 1, 7) \) or \( (-2, 5, 3) \).