JEE Main 2022 · 28 Jun · Shift 2 · Mathematics Q16

MathematicsSingle correct+4 / -1Question 16 of 90

Let the plane $ax+by+cz=d$ pass through $(2,3,-5)$ and is perpendicular to the planes $2x+y-5z=10$ and $3x+5y-7z=12$ .

If $a,b,c,d$ are integers $d>0$ and $\gcd(|a|,|b|,|c|,d)=1$ , then the value of $a+7b+c+20d$ is equal to :

Answer and worked solution

Official answer: D

Normal. $(2,1,-5)\times(3,5,-7)=(18,-1,7)$ ; through $(2,3,-5)$ : $18x-y+7z=-2$ .

Normalise with $d>0$ : $-18x+y-7z=2$ .

Value. $-18+7-7+40=22$ .

(D)

Topics:Equation of a plane Cross product

JEE Main 2022 · 28 Jun · Shift 2, Mathematics question 16