Question

$\frac{x-2}{3} = \frac{y-1}{-5} = \frac{z+2}{2}$ lies in the plane $x+3y-\alpha z + \beta = 0$, then value of $\alpha \beta$ is

• A. 42
• B. 1
• C. -42
• D. -2

Answer 1

Answer: (C)

-42

Answer 2

Solution:

The line is given by

$$\frac{x-2}{3} = \frac{y-1}{-5} = \frac{z+2}{2}.$$

Parametrize it by $t$:

$$x = 2 + 3t,\quad y = 1 - 5t,\quad z = -2 + 2t.$$

For the line to lie in the plane

$$x+3y-\alpha z+\beta=0,$$

the following two conditions must hold:

1. Direction vector condition:
   The direction vector of the line is $\mathbf{d} = (3,-5,2)$ and the normal to the plane is $\mathbf{n} = (1,3,-\alpha)$. Since the line lies in the plane,

   $$\mathbf{n} \cdot \mathbf{d}=0 \quad \Rightarrow \quad 1\cdot3 + 3\cdot(-5) + (-\alpha)\cdot2 = 0.$$

   Simplify:

   $$3 - 15 - 2\alpha = 0 \quad \Rightarrow \quad -12 - 2\alpha = 0 \quad \Rightarrow \quad \alpha = -6.$$

2. Point condition:
   The line passes through the point $(2,1,-2)$ (when $t=0$). Substitute into the plane:

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

   Calculate:

   $$2 + 3 - 12 + \beta = 0 \quad \Rightarrow \quad -7 + \beta = 0 \quad \Rightarrow \quad \beta = 7.$$

The product is:

$$\alpha\beta = (-6)(7) = -42.$$