Answer
Normal form of the plane:$\begin{bmatrix}{2 \\5\\0}\end{bmatrix} \cdot \begin{bmatrix}{x-3 \\y\\z+2}\end{bmatrix}=0$;
General form of the equation of the plane: $2x+5y=6$
Work Step by Step
The normal form of a plane is: $n \cdot (x-p)=0$
and $n=\begin{bmatrix}{a \\b\\c}\end{bmatrix}$
Here: $a=2, b=5,c=0$
Now, the normal form of the plane is:$\begin{bmatrix}{2 \\5\\0}\end{bmatrix} \cdot \begin{bmatrix}{x-3 \\y\\z+2}\end{bmatrix}=0$
The general form of equation of a plane is: $ax+by+cz=d$
Thus, the General form of the equation of a plane is: $2x+5y=d$
or, $2(3)+5(0)+0=d \implies d=6$
Hence, the normal form of the plane is: $\begin{bmatrix}{2 \\5\\0}\end{bmatrix} \cdot \begin{bmatrix}{x-3 \\y\\z+2}\end{bmatrix}=0$;
Thus, the General form of the equation of a plane is: $2x+5y=6$
