Let the line x - 2 3 = y - 1 - 5 = z + 2 2 lie in the plane x + 3 y - z + = 0, then equals:

Question

Let the line $\frac{x - 2}{3} = \frac{y - 1}{- 5} = \frac{z + 2}{2}$ lie in the plane $x + 3 y - α z + β = 0,$ then $(α, β)$ equals:

Smart Achievers · Accepted Answer

(1)

Ans. $(- 6, 7)$

Explanation:

Since the line $\frac{x - 2}{3} = \frac{y - 1}{- 5} = \frac{z + 2}{2}$ lies in the plane

$x + 3 y - α z + β = 0$

Therefore, the point $(2, 1, - 2)$ lies on the plane.

i.e., $2 + 3 + 2 α + β = 0$

or $2 α + β + 5 = 0$ $\dots (i)$

Also, the normal to the plane will be perpendicular to the line.

Therefore,

$3 \times 1 - 5 \times 3 + 2 \times (- α) = 0$

$\Rightarrow$ $α = - 6$

From equation (i), we have

$β = 7$

Therefore, $(α, β) = (- 6, 7)$

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