Let the line `(x-2)/(3)=(y-1)/(-5)=(z+2)/(2)` lies in the plane `x+3y-alpha z +beta=0`. Then, `(alpha,beta)` equals

Options

A. `(6,-17)`
B. `(-6,7)`
C. `(5,-15)`
D. `(-5,15)`

Explanation

To solve the problem, we need to find the values of \( \alpha \) and \( \beta \) such that the line given by the equation

\[ \frac{x-2}{3} = \frac{y-1}{-5} = \frac{z+2}{2} \]

lies in the plane defined by

\[ x + 3y - \alpha z + \beta = 0. \]

### Step 1: Identify the direction ratios of the line and the normal vector of the plane.

The line can be expressed in parametric form as:
- \( x = 3t + 2 \)
- \( y = -5t + 1 \)
- \( z = 2t - 2 \)

From this, we can identify the direction ratios of the line as \( (3, -5, 2) \).

The normal vector of the plane \( x + 3y - \alpha z + \beta = 0 \) is given by the coefficients of \( x, y, z \), which are \( (1, 3, -\alpha) \).

### Step 2: Use the condition for perpendicularity.

Since the line lies in the plane, the direction ratios of the line must be perpendicular to the normal vector of the plane. This means that their scalar product must be zero:

\[ (1, 3, -\alpha) \cdot (3, -5, 2) = 0. \]

Calculating the scalar product:

\[ 1 \cdot 3 + 3 \cdot (-5) + (-\alpha) \cdot 2 = 0. \]

This simplifies to:

\[ 3 - 15 - 2\alpha = 0. \]

### Step 3: Solve for \( \alpha \).

Rearranging the equation gives:

\[ -12 - 2\alpha = 0 \implies 2\alpha = -12 \implies \alpha = -6. \]

### Step 4: Find a point on the line.

Next, we need a point on the line to find \( \beta \). We can take \( t = 0 \):

\[ x = 2, \quad y = 1, \quad z = -2. \]

So the point \( (2, 1, -2) \) lies on the line.

### Step 5: Substitute the point into the plane equation.

Now, substitute \( (x, y, z) = (2, 1, -2) \) into the plane equation:

\[ 2 + 3(1) - (-6)(-2) + \beta = 0. \]

This simplifies to:

\[ 2 + 3 - 12 + \beta = 0 \implies -7 + \beta = 0 \implies \beta = 7. \]

### Final Answer

Thus, the values of \( \alpha \) and \( \beta \) are:

\[ (\alpha, \beta) = (-6, 7). \]