MHB Finding equation normal to a plane and certain point

brunette15
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For the following question I am given a plane: x + 2y + 3z = 12. I want to find the equation of a line normal to the plane and going through the point (4,6,8). I am trying to use the formula N . (r - r0) = 0 however seem to be getting the incorrect answer :(
 
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The line equation that passes through $a$ and is parallel to $b$ is $l(t)=a+tb$.

From the equation of the plane $x + 2y + 3z = 12$, we have that a normal vector to the plane is $(1,2,3)$.

We are looking for a line that is normal to the plane, so parallel to the vector $(1,2,3)$ and passes through the point $(4,6,8)$.

Can you find now the line equation?
 
evinda said:
The line equation that passes through $a$ and is parallel to $b$ is $l(t)=a+tb$.

From the equation of the plane $x + 2y + 3z = 12$, we have that a normal vector to the plane is $(1,2,3)$.

We are looking for a line that is normal to the plane, so parallel to the vector $(1,2,3)$ and passes through the point $(4,6,8)$.

Can you find now the line equation?

Thankyou so much I was able to figure it out from there :)
 

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