JEE Main 2022 · 25 Jul · Shift 2 · Mathematics Q15

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MathematicsSingle correct+4 / -1Question 15 of 90

A plane E is perpendicular to the two planes $2x-2y+z=0$ and $x-y+2z=4$ , and passes through the point $P(1,-1,1)$ . If the distance of the plane E from the point $Q(a,a,2)$ is $3\sqrt2$ , then $(PQ)^2$ is equal to

Answer and worked solution

Official answer: C

Plane E. Normal $(2,-2,1)\times(1,-1,2)\parallel(1,1,0)$ , through P: $x+y=0$ .

Distance $\frac{|2a|}{\sqrt2}=3\sqrt2$ , $a^2=9$ . $PQ^2$ $=2a^2+3=21$ .

(C)

Topics:Equation of a plane Angles and distances

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JEE Main 2022 · 25 Jul · Shift 2, Mathematics question 15