If the points -x- - - - -y- - - - -z and -1- -1- -1 are coplanar - R then

The correct option is D #10; 1x+1y+1z=1 #10; V⃗= (x, y,0) U⃗= (0,y, z) W⃗= ( 1, 1,z 1) 0=x amp; y #160; amp;0 0 #1 ...

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Addition of Vectors

If the points...

Question

If the points $(α - x, α, α)$ , $(α, α - y, α)$ , $(α, α, α - z)$ and $(α - 1, α - 1, α - 1)$ are coplanar $α \in R$ then

x y + y z + z x = 1

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\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1

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\frac{1}{1 - x} + \frac{1}{1 - y} + \frac{1}{1 - z} = 1

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x y z = 1

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α x - y - z = α - 1, x - α y - z = α - 1, x - y - α z = α - 1

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α x + y + z = α - 1, x + α y + z = α - 1, x + y + α z = α - 1

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\begin{matrix} α x + y + z = α - 1 x + α y + z = α - 1 x + y + α z = α - 1 \end{matrix}

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α x + y + z = α - 1, x + α y + z = α - 1, x + y + α z = α - 1

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α \in R,

L_{1} : \frac{x + 1}{2} = \frac{y - 2}{- 1} = \frac{z - 1}{1}

L_{2} : \frac{x + 2}{α} = \frac{y + 1}{5 - α} = \frac{z + 1}{1}

L_{2}

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