The equation of plane passes through (1,2,3) and perpendicular to the line x2=y3=z1 is
A
x+2y+3z=14
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B
x−2y−3z+12=0
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C
3x+2y+z=8
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D
2x+3y+z=11
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Solution
The correct option is D2x+3y+z=11 As, the line x2=y3=z1 is perpendicular to the plane
So, D.r′s os normal =(a,b,c)=(2,3,1)
So, equation is : a(x−x1)+b(y−y1)+c(z−z1)=0 ⇒2(x−1)+3(y−2)+1(z−3)=0⇒2x+3y+z=11