The equation of a plane passing through the line of intersection of the planes x + 2y + 3z = 2 and x – y + z = 3 and at a distance \frac{2}{\sqrt{3}} from the point (3, 1,

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Mathematics

Plane and Its Different Forms

Question

The equation of a plane passing through the line of intersection of the planes x + 2y + 3z = 2 and
x – y + z = 3 and at a distance $\frac{2}{\sqrt{3}}$ from the point (3, 1, – 1) is

5x – 11y + z = 17

$x + y + z = \sqrt{3}$

$x - \sqrt{2} y = 1 - \sqrt{2}$

$\sqrt{2} x + y = 3 \sqrt{2} - 1$

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Let the plane passes through the line of intersection of given place is

(x + 2y + 3z – 2) + l (x – y + z – 3) = 0

$\Rightarrow$ x (1 + $λ$ ) + y (2 – $λ$ ) + z (3 + $λ$ ) – (2 + 3 $λ$ ) = 0 .......(1)

Its distance from (3, 1, –1) is $\frac{2}{\sqrt{3}}$ .

$\Rightarrow \frac{3 (1 + λ) + (2 - λ) - (3 + λ) - (2 + 3 λ)}{\sqrt{{(1 + λ)}^{2} + {(2 - λ)}^{2} + {(3 + λ)}^{2}}} = \pm \frac{2}{\sqrt{3}}$

$\Rightarrow 3 λ^{2} - {(1 + λ)}^{2} + {(2 - λ)}^{2} + {(3 + λ)}^{2} \Rightarrow λ = \frac{- 7}{2}$

Put value of l in equation (1) we get 5x – 11y + z = 17.

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