Question

Let the line \(\frac{x-2}{3} = \frac{y-1}{-5} = \frac{z+2}{2}\) lie in the plane \(x + 3y - \alpha z + \beta = 0\), then the value of \((\beta - \alpha)\) is equal to

• A. \(1\)
• B. \(13\)
• C. \(7\)
• D. \(-6\)

Hint:

For a line lying in a plane, always use both conditions:
• direction vector \(\perp\) plane normal
• point on line satisfies plane equation e}

Answer 1

The Correct Option is B

Concept:
If a line lies in a plane, then:
• the direction vector of the line must be perpendicular to the normal vector of the plane,
• and any point on the line must satisfy the plane equation e} ip

Step 1: Write a point and direction vector of the line.
From \[ \frac{x-2}{3} = \frac{y-1}{-5} = \frac{z+2}{2} \] a point on the line is: \[ (2,1,-2) \] and its direction vector is: \[ (3,-5,2) \] ip

Step 2: Use perpendicularity with the plane normal.
The plane is: \[ x+3y-\alpha z+\beta=0 \] So its normal vector is: \[ (1,3,-\alpha) \] For the line to lie in the plane: \[ (3,-5,2)\cdot(1,3,-\alpha)=0 \] \[ 3-15-2\alpha=0 \] \[ -12-2\alpha=0 \] \[ \alpha=-6 \] ip

Step 3: Use the point condition to find \(\beta\).
Substitute \((2,1,-2)\) in the plane: \[ 2+3(1)-(-6)(-2)+\beta=0 \] \[ 2+3-12+\beta=0 \] \[ -7+\beta=0 \] \[ \beta=7 \] ip

Step 4: Find \(\beta-\alpha\).
\[ \beta-\alpha = 7-(-6)=13 \] ip Hence, the correct answer is:
\[ \boxed{(B)\ 13} \]