Let the plane x + 3 y - 2 z + 6 = 0 meet the coordinate axes at the points A, B, C. If the orthocenter of the triangle ABC is ( 6 7 ) , then 98 2 is equal to _____ . [2023]

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Solution

Q.
Let the plane $x + 3 y - 2 z + 6 = 0$ meet the coordinate axes at the points A, B, C. If the orthocenter of the triangle ABC is $(α, β, \frac{6}{7})$ , then $98 {(α + β)}^{2}$ is equal to _____ . [2023]

Ans.
(288)

Plane $x + 3 y - 2 z + 6 = 0$ meets the coordinate axes at the points A, B, C.

$For x-axis: x + 0 - 0 + 6 = 0 \Rightarrow x = - 6 i.e., (- 6, 0, 0)$

$For y-axis: 0 + 3 y - 0 + 6 = 0 \Rightarrow y = - 2 i.e., (0, - 2, 0)$

$For z-axis: 0 + 0 - 2 z + 6 = 0 \Rightarrow z = 3 i.e., (0, 0, 3)$

$\vec{A B} = 6 \hat{i} - 2 \hat{j}, \vec{B C} = 2 \hat{j} + 3 \hat{k}, \vec{A C} = 6 \hat{i} + 3 \hat{k}$

Now, $\vec{A P} \cdot \vec{B C} = 0$

$∴$ $(α + 6, β, \frac{6}{7}) \cdot (0, 2, 3) = 0$

$\Rightarrow 2 β + 3 \cdot \frac{6}{7} = 0 \Rightarrow β = \frac{- 9}{7}$

Similarly $\vec{C P} \cdot \vec{A B} = 0$

$(α, β, - \frac{15}{7}) \cdot (6, - 2, 0) = 0 \Rightarrow 6 α - 2 β = 0$

$\Rightarrow 6 α - 2 (\frac{- 9}{7}) = 0 \Rightarrow 6 α = \frac{- 18}{7} \Rightarrow α = \frac{- 3}{7}$

$∴$ $98 {(α + β)}^{2} = 98 {(\frac{- 3}{7} - \frac{9}{7})}^{2} = 98 {(\frac{- 12}{7})}^{2} = 98 \cdot \frac{144}{49} = 288$

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