If the line `(x-2)/(3)=(y+1)/(2)=(z-1)/(-1)` intersects the plane `2x+3y-z+13=0` at a point P and plane `3x+y+4z=16` at a point Q then PQ is equal to (A) `14` (B) `sqrt(14)` (C) `2sqrt(7)` (D) `2sqrt(14)`

Options

A. `14`
B. `sqrt(14)`
C. `2sqrt(7)`
D. `2sqrt(14)`

Answer

To solve the problem step by step, we will find the points of intersection of the given line with the two planes and then calculate the distance between these two points.

### Step 1: Parametrize the Line
The line is given by the equation:
\[ \frac{x-2}{3} = \frac{y+1}{2} = \frac{z-1}{-1} \]
Let this common ratio be \( \lambda \). Then we can express the coordinates \( x, y, z \) in terms of \( \lambda \):
\[ x = 3\lambda + 2, \quad y = 2\lambda - 1, \quad z = -\lambda + 1 \]

### Step 2: Find Point P (Intersection with the First Plane)
The first plane is given by the equation:
\[ 2x + 3y - z + 13 = 0 \]
Substituting the parametric equations of the line into the plane equation:
\[ 2(3\lambda + 2) + 3(2\lambda - 1) - (-\lambda + 1) + 13 = 0 \]
Expanding this:
\[ 6\lambda + 4 + 6\lambda - 3 + \lambda - 1 + 13 = 0 \]
Combining like terms:
\[ (6\lambda + 6\lambda + \lambda) + (4 - 3 - 1 + 13) = 0 \]
\[ 13\lambda + 13 = 0 \]
Solving for \( \lambda \):
\[ \lambda = -1 \]

### Step 3: Calculate Coordinates of Point P
Substituting \( \lambda = -1 \) back into the parametric equations:
\[ x_P = 3(-1) + 2 = -1, \quad y_P = 2(-1) - 1 = -3, \quad z_P = -(-1) + 1 = 2 \]
Thus, the coordinates of point P are:
\[ P(-1, -3, 2) \]

### Step 4: Find Point Q (Intersection with the Second Plane)
The second plane is given by:
\[ 3x + y + 4z = 16 \]
Substituting the parametric equations into this plane equation:
\[ 3(3\lambda + 2) + (2\lambda - 1) + 4(-\lambda + 1) = 16 \]
Expanding this:
\[ 9\lambda + 6 + 2\lambda - 1 - 4\lambda + 4 = 16 \]
Combining like terms:
\[ (9\lambda + 2\lambda - 4\lambda) + (6 - 1 + 4) = 16 \]
\[ 7\lambda + 9 = 16 \]
Solving for \( \lambda \):
\[ 7\lambda = 7 \implies \lambda = 1 \]

### Step 5: Calculate Coordinates of Point Q
Substituting \( \lambda = 1 \) back into the parametric equations:
\[ x_Q = 3(1) + 2 = 5, \quad y_Q = 2(1) - 1 = 1, \quad z_Q = -(1) + 1 = 0 \]
Thus, the coordinates of point Q are:
\[ Q(5, 1, 0) \]

### Step 6: Calculate Distance PQ
The distance \( PQ \) is given by the formula:
\[ PQ = \sqrt{(x_Q - x_P)^2 + (y_Q - y_P)^2 + (z_Q - z_P)^2} \]
Substituting the coordinates of points P and Q:
\[ PQ = \sqrt{(5 - (-1))^2 + (1 - (-3))^2 + (0 - 2)^2} \]
Calculating each term:
\[ = \sqrt{(5 + 1)^2 + (1 + 3)^2 + (0 - 2)^2} \]
\[ = \sqrt{6^2 + 4^2 + (-2)^2} \]
\[ = \sqrt{36 + 16 + 4} \]
\[ = \sqrt{56} = \sqrt{4 \times 14} = 2\sqrt{14} \]

### Final Answer
Thus, the distance \( PQ \) is:
\[ \boxed{2\sqrt{14}} \]