\[P\]

? 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

\[2x-3y+z=-4\]

.