The line x-2/3 - y-1/-5 = z+2/2 lies in the plane x + 3y - α z + β = 0, then value of α β is

Correct Answer is : −42-42−42

Correct Answer is : −42-42−42

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MHT CET 2021 Math Solved Paper

Question : 74 of 100

Marks: +1, -0

The line

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

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

\alpha \beta

Solution:

Given equation of line

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

. . . (i)

The direction ratios of the normal are

(1,3,-\alpha)

The direction ratios of the line are

(3,-5,2)

and equation of given plane

x + 3y - a z + \beta = 0

. . . (ii)

Four lines are perpendicular

\Rightarrow a_1 a_2 + b_1 b_2 + c_1 c_2 = 0

\Rightarrow 3 - 15 + 2 \alpha = 0

\Rightarrow 2 \alpha = -12 \Rightarrow \alpha = -6

(2,1,-2)

lies on the plane, so

2 + 3 + 6(-2) + \beta = 0 \Rightarrow \beta = 7

\alpha \cdot \beta = -6 \times 7 = -42

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