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# Find the equation of the plane passing through the point (–1, 3, 2) and perpen- dicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0

SHAIK AASIF AHAMED
7 years ago
equation of a plane passing through (-1,3,2) is
a(x+1)+b(y-3)+c(z-2)=0
since the plane is perpendicular to given 2 planes
a+2b+3c=0;
9a+9b+3c=0
by solving we get
$\frac{a}{6-18}=\frac{b}{3-27}=\frac{c}{9-18}$
$\frac{a}{4}=\frac{b}{8}=\frac{c}{3}$
the equation of plane is 4(x+1)+8(y-3)+3(z-2)=0
4x+8y+3z-26=0
Thanks & Regards
Jitender Singh
IIT Delhi