Let the linex-23 - y-1-5 - z-22lies in the plane x - 3y - - z - - - 0- Then - - equals

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Condition for a Line to Lie on a Plane

Let the linex...

Question

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

$(6, - 17)$

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$(- 6, 7)$

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$(5, - 15)$

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$(- 5, 15)$

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Solution

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Similar questions

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

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

x + 3 y - α z + β = 0

(α, β)

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

x + 3 y - α z + β = 0

(α, β)

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

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

x + 3 y - α z + β = 0.

(α, β)

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