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Question

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
x2y3z+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 D 2x+3y+z=11
As, the line x2=y3=z1 is perpendicular to the plane
So, D.rs os normal =(a,b,c)=(2,3,1)
So, equation is : a(xx1)+b(yy1)+c(zz1)=0
2(x1)+3(y2)+1(z3)=02x+3y+z=11

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