Equation of a Plane Parallel to a Given Plane

Suppose a plane  
\[P\]
? passes through the point  
\[(1,3,4)\]
  and is parallel to the plane  
\[2x-3y+z=7\]
.
How do we find the equation of  
\[P\]
?
Because the planes are parallel they have the same normal. The normal to a plane given in Cartesian form is equal to the coefficients of the equation of the plane, written as a vector, in this case  
\[\mathbf{n}= \begin{pmatrix}2\\-3\\\end{pmatrix}\]
, so both have the same coefficients in the Cartesian form of the plane equation, and the plane  
\[P\]
  will have equation  
\[2x-3y+z=d\]
.  
\[d\]
  can be found by substituting the point above into the equation.
\[d=1(1)-3(3)+4=-4\]
.
The equation of the plane is  
\[2x-3y+z=-4\]
.

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