JEE Main 2022 · 29 Jun · Shift 2 · Mathematics Q12

← JEE Main 2022 · 29 Jun · Shift 2

MathematicsSingle correct+4 / -1Question 12 of 90

Let $\dfrac{x-2}3=\dfrac{y+1}{-2}=\dfrac{z+3}{-1}$ lie on the plane $px-qy+z=5$ , for some $p,q\in\mathbb R$ . The shortest distance of the plane from the origin is:

\sqrt{\dfrac3{109}}

\sqrt{\dfrac5{142}}

\dfrac5{\sqrt{71}}

\dfrac1{\sqrt{142}}

Answer and worked solution

Official answer: B

Conditions. The point $(2,-1,-3)$ is on the plane: $2p+q=8$ . The direction $(3,-2,-1)$ is perpendicular to $(p,-q,1)$ : $3p+2q=1$ . So $p=15$ , $q=-22$ .

Plane $15x+22y+z=5$ : distance $\dfrac5{\sqrt{710}}=\sqrt{\dfrac5{142}}$ .

(B)

Topics:Equation of a plane Angles and distances

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JEE Main 2022 · 29 Jun · Shift 2, Mathematics question 12