# Find the vector and cartesian equations of the plane through the point $(2 , -1 , 4 )$ and parallel to the plane $\overrightarrow{r} . (\overrightarrow{4i}-\overrightarrow{12j}-\overrightarrow{3k})=7.$

Toolbox:
• Equation of the plane through a point $A(x_1, y_1, z_1) [pv=\overrightarrow a]$ and perpendicular to vector $\overrightarrow n = a \overrightarrow i+b \overrightarrow j+c \overrightarrow k$
• $\overrightarrow r. \overrightarrow n = \overrightarrow a. \overrightarrow n$ (vector equation )
• $a(x-x_1)+b(y-y_1)+c(z-z_1)=0$ (cartesian equation)
Step 1
The required plane passes through the point $A(2, -1, 4)$ i.e., $\overrightarrow a = 2\overrightarrow i-\overrightarrow j+4\overrightarrow k$
It is parallel to the plane $\overrightarrow r.(4\overrightarrow i-12\overrightarrow j-3\overrightarrow k)=7$
The normal vectors to this plane is $\overrightarrow n=4\overrightarrow i-12\overrightarrow j-3\overrightarrow k.\: \therefore$ it is normal to the required plane.
Step 2
The vector equation of the required plane is $\overrightarrow r.\overrightarrow n=\overrightarrow a.\overrightarrow n$
$\overrightarrow r.(4\overrightarrow i-12\overrightarrow j-3\overrightarrow k)=(2\overrightarrow i-\overrightarrow j+4\overrightarrow k).(4\overrightarrow i-12\overrightarrow j-3\overrightarrow k)$
$\overrightarrow r.(4\overrightarrow i-12\overrightarrow j-3\overrightarrow k)=8+12-12=8$
$\overrightarrow r.(4\overrightarrow i-12\overrightarrow j-3\overrightarrow k)=8$
$\overrightarrow r=x\overrightarrow i+y\overrightarrow j+z\overrightarrow k$
Step 3
The cartesian equation is $4x-12y-3z=8$