Let be the point of intersection of the line and the plane . If the distance of the point from the plane is , then and are the roots of the equation:

Explanation: To find point , let . Point lies on . Point . Distance . and . Equation having roots and is .

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JEE 2023 Mathematics PYQ — Let be the point of intersection of the line and the plane . If t… | Mathem Solvex | Mathem Solvex

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JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $P$ be the point of intersection of the line $\frac{x+3}{3} = \frac{y+2}{1} = \frac{1-z}{2}$ and the plane $x + y + z = 2$ . If the distance of the point $P$ from the plane $3x - 4y + 12z = 32$ is $q$ , then $q$ and $2q$ are the roots of the equation:

Choose the correct answer:

A.
$(1) x^2 + 18x - 72 = 0 \quad (2) x^2 + 18x + 72 = 0$
B.
$(1) x^2 + 18x - 72 = 0 \quad (2) x^2 + 18x + 72 = 0$

Correct Answer:

$(3) x^2 - 18x - 72 = 0 \quad (4) x^2 - 18x + 72 = 0$

Explanation

To find point $P$ , let $\frac{x+3}{3} = \frac{y+2}{1} = \frac{1-z}{2} = k \Rightarrow x = 3k-3, y = k-2, z = 1-2k$ .

Point $P$ lies on $x+y+z=2 \Rightarrow (3k-3) + (k-2) + (1-2k) = 2 \Rightarrow 2k = 6 \Rightarrow k=3$ .

Point $P = (6, 1, -5)$ .

Distance $q = \frac{|3(6) - 4(1) + 12(-5) - 32|}{\sqrt{3^2 + (-4)^2 + 12^2}} = \frac{|18 - 4 - 60 - 32|}{\sqrt{169}} = \frac{78}{13} = 6$ .

$q = 6$ and $2q = 12$ .

Equation having roots $q$ and $2q$ is $x^2 - (6+12)x + (6 \times 12) = 0 \Rightarrow x^2 - 18x + 72 = 0$ .

Explanation

To find point $P$ , let $\frac{x+3}{3} = \frac{y+2}{1} = \frac{1-z}{2} = k \Rightarrow x = 3k-3, y = k-2, z = 1-2k$ .

Point $P$ lies on $x+y+z=2 \Rightarrow (3k-3) + (k-2) + (1-2k) = 2 \Rightarrow 2k = 6 \Rightarrow k=3$ .

Point $P = (6, 1, -5)$ .

Distance $q = \frac{|3(6) - 4(1) + 12(-5) - 32|}{\sqrt{3^2 + (-4)^2 + 12^2}} = \frac{|18 - 4 - 60 - 32|}{\sqrt{169}} = \frac{78}{13} = 6$ .

$q = 6$ and $2q = 12$ .

Equation having roots $q$ and $2q$ is $x^2 - (6+12)x + (6 \times 12) = 0 \Rightarrow x^2 - 18x + 72 = 0$ .

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C.
$(3) x^2 - 18x - 72 = 0 \quad (4) x^2 - 18x + 72 = 0$

D.
$(3) x^2 - 18x - 72 = 0 \quad (4) x^2 - 18x + 72 = 0$
(Correct Answer)

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