41. Let the line x-2/3=y-1/-5=z+2/2 lie in the plane x+3y-az+b=0 then (a,b) is

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\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)

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)

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