The distance of the point P `(4, 6 – 2)` from the line passing through the point (–3, 2, 3) and parallel to a line with direction ratios 3, 3, –1 is equal to:

Options

A. `2sqrt3`
B. `sqrt14`
C. 3
D. `sqrt6`

Answer

To find the distance of the point \( P(4, 6, -2) \) from the line passing through the point \( (-3, 2, 3) \) and parallel to the line with direction ratios \( (3, 3, -1) \), we can follow these steps:

### Step 1: Write the equation of the line
The line can be represented in parametric form using the point \( (-3, 2, 3) \) and the direction ratios \( (3, 3, -1) \). The parametric equations of the line can be written as:
\[ x = -3 + 3t, \quad y = 2 + 3t, \quad z = 3 - t \]
where \( t \) is a parameter.

### Step 2: Find a general point on the line
Let \( Q \) be a point on the line. From the parametric equations, we can express the coordinates of point \( Q \) as:
\[ Q(t) = (-3 + 3t, 2 + 3t, 3 - t) \]

### Step 3: Find the direction ratios of line \( PQ \)
The direction ratios of the line segment \( PQ \) from point \( P(4, 6, -2) \) to point \( Q(t) \) can be calculated as:
\[ PQ = Q(t) - P = \left((-3 + 3t - 4), (2 + 3t - 6), (3 - t + 2)\right) \]
This simplifies to:
\[ PQ = (3t - 7, 3t - 4, -t + 5) \]

### Step 4: Set up the condition for perpendicularity
Since \( PQ \) is perpendicular to the line, the dot product of the direction ratios of \( PQ \) and the direction ratios of the line must equal zero:
\[ (3t - 7) \cdot 3 + (3t - 4) \cdot 3 + (-t + 5) \cdot (-1) = 0 \]
Expanding this gives:
\[ 9t - 21 + 9t - 12 + t - 5 = 0 \]
Combining like terms results in:
\[ 19t - 38 = 0 \]

### Step 5: Solve for \( t \)
Solving for \( t \):
\[ 19t = 38 \implies t = 2 \]

### Step 6: Find the coordinates of point \( Q \)
Substituting \( t = 2 \) back into the parametric equations gives:
\[ Q(2) = (-3 + 3 \cdot 2, 2 + 3 \cdot 2, 3 - 2) = (3, 8, 1) \]

### Step 7: Calculate the distance \( d \) from point \( P \) to point \( Q \)
The distance \( d \) can be calculated using the distance formula:
\[ d = \sqrt{(4 - 3)^2 + (6 - 8)^2 + (-2 - 1)^2} \]
Calculating each term:
\[ = \sqrt{(1)^2 + (-2)^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]

### Final Answer
Thus, the distance of the point \( P(4, 6, -2) \) from the line is \( \sqrt{14} \).