Answer
$x+4y-3z=6$
Work Step by Step
Let $(x_{0},y_{0},z_{0})$ be a point on the plane and $\lt a,b,c\gt$ be a normal vector to the plane.
Then a vector equation of the plane is $\lt a,b,c\gt$.$(x_{0},y_{0},z_{0})=0$
A scalar equation is: $a(x-x_{0}),b(y-y_{0}),c(z-z_{0})$
The plane $(2,1,0)$ and has a normal vector $=\lt 1,4,-3\gt$ (the same normal vector as its parallel plane $x+4y-3z=1$,
Therefore,
$\lt 1,4,-3\gt$ $(x-2,y-1,z)=0$
A scalar equation is:
$x-2+4y-4-3z=0$
Hence, the required equation of the plane is $x+4y-3z=6$
