Let P be the plane passing through the line x−11=y−2−3=z+57 and the point (2, 4, -3). if the image of the point (-1, 3, 4) in the plane P is (α,β,γ) then α+β+γ is equal to

The equation of plane passing through the points (1, 2, -5) and (2, 4, -3) and parallel the line having Drs (1,-3, 7) isx−1y−2z+52−14+2−3+51−37=0⇒x−1y−2z+51221−37=0⇒4x−y−z−7=0Let (α,β,γ) be the image of the point (-1, 3, 4) in the plane Then α+14=β−3−1=γ−4−1=−2(−4−3−4−7)42+(−1)2+(−1)2=2⇒α=7,β=1,γ=2⇒α+β+γ=10

Let P be the plane passing through the line x−11=y−2−3=z+57 and the point (2, 4, -3). if the image of the point (-1, 3, 4) in the plane P is (α,β,γ) then α+β+γ is equal to

The equation of plane passing through the points (1, 2, -5) and (2, 4, -3) and parallel the line having Drs (1,-3, 7) isx−1y−2z+52−14+2−3+51−37=0⇒x−1y−2z+51221−37=0⇒4x−y−z−7=0Let (α,β,γ) be the image of the point (-1, 3, 4) in the plane Then α+14=β−3−1=γ−4−1=−2(−4−3−4−7)42+(−1)2+(−1)2=2⇒α=7,β=1,γ=2⇒α+β+γ=10

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Let P be the plane passing through the line $\frac{x - 1}{1} = \frac{y - 2}{- 3} = \frac{z + 5}{7}$ and the point (2, 4, -3). if the image of the point
(-1, 3, 4) in the plane P is $(α, β, γ) then α + β + γ$ is equal to

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Detailed Solution

The equation of plane passing through the points (1, 2, -5) and (2, 4, -3) and parallel the line having Drs (1,-3, 7) is
$\begin{array}{l} |\begin{array}{ccc} x - 1 & y - 2 & z + 5 \\ 2 - 1 & 4 + 2 & - 3 + 5 \\ 1 & - 3 & 7 \end{array}| = 0 \Rightarrow |\begin{array}{ccc} x - 1 & y - 2 & z + 5 \\ 1 & 2 & 2 \\ 1 & - 3 & 7 \end{array}| = 0 \\ \Rightarrow 4 x - y - z - 7 = 0 \end{array}$
Let $(α, β, γ)$ be the image of the point (-1, 3, 4) in the plane Then
$\begin{array}{l} \frac{α + 1}{4} = \frac{β - 3}{- 1} = \frac{γ - 4}{- 1} = \frac{- 2 (- 4 - 3 - 4 - 7)}{4^{2} + (- 1)^{2} + (- 1)^{2}} = 2 \\ \Rightarrow α = 7, β = 1, γ = 2 \Rightarrow α + β + γ = 10 \end{array}$