Let the plane $\mathrm{x}+3 \mathrm{y}-2 \mathrm{z}+6=0$ meet the co-ordinate axes at the points $\mathrm{A}, \mathrm{B}, \mathrm{C}$. If the orthocentre of the triangle ABC is $\left(\alpha, \beta, \frac{6}{7}\right)$, then $98(\alpha+\beta)^{2}$ is equal to

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Three Dimensional Geometry

Let the plane

x + 3 y - 2 z + 6 = 0

meet the co-ordinate axes at the points

A, B, C

. If the orthocentre of the triangle ABC is

(α, β, \frac{6}{7})

, then

98 (α + β)^{2}

is equal to

Answer: 288

Solution

The plane

x + 3 y - 2 z + 6 = 0

meets the axes at

A (- 6, 0, 0)

B (0, - 2, 0)

C (0, 0, 3)

. Let the orthocentre be

H = (α, β, \frac{6}{7})

. The condition

A H ⊥ B C

gives

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

, so

2 β + \frac{18}{7} = 0

, giving

β = - \frac{9}{7}

. The condition

C H ⊥ A B

gives

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

, so

6 α - 2 β = 0

, giving

α = \frac{β}{3} = - \frac{3}{7}

. Thus

α + β = - \frac{12}{7}

, and

98 (α + β)^{2} = 98 \cdot \frac{144}{49} = 288

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Integer answer

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