JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let P₁ be the plane 3x – y – 7z = 11 and P₂ be the plane passing through the points (2, –1, 0), (2, 0, –1), and (5, 1, 1). If the foot of the perpendicular drawn from the point (7, 4, –1) on the line of intersection of the planes P₁ and P₂ is (α, β, γ), then α + β + γ is equal to _____.

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Explanation

$\Rightarrow x - y - z = 3 \dots (i)$

Now, direction ratios of line of intersection of P₁ and P₂is
$ \hat{i} 1 3 am p; \hat{j} am p; - 1 am p; - 1 am p; \hat{k} am p; - 1 am p; - 7 $

$= \hat{i} (7 - 1) - \hat{j} (- 7 + 3) + \hat{k} (- 1 + 3)$
$= 6 \hat{i} + 4 \hat{j} + 2 \hat{k}$

At z = 0, x - y = 3 [from (i)]
$3 x - y = 11$

on solving, we get
$x = 4 and y = 1$

So, equation of line is
$\frac{x - 4}{6} = \frac{y - 1}{4} = \frac{z}{2}$

$\frac{x - 4}{6} = \frac{y - 1}{4} = \frac{z - 2}{6} = k$

$∴ (α, β, γ) = (6 k + 4, 4 k + 1, 2 k)$
$\Rightarrow (6) (α - 7) + 4 (β - 4) + 2 (γ + 1) = 0$
$\Rightarrow 6 (6 k + 4 - 7) + 4 (4 k + 1 - 4) + 2 (2 k + 1) = 0$
$\Rightarrow 36 k - 18 + 16 k - 12 + 4 k + 4 = 0$
$\Rightarrow 56 k = 26 \Rightarrow k = \frac{1}{2}$

\text{So, } $α = 7, β = 3 and γ = 1$
$∴ α + β + γ = 7 + 3 + 1 = 11$

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