Write the perpendicular distance of the plane $$\overrightarrow r.(2\hat i - \hat j + 2\hat k)=12$$ from the origin.

Toolbox:
• The equation of the form $lx+my+nz=d$ where $l,m,n$ are the direction cosines of the normal to the plane,and $d$ is the distance of the normal from the origin.
Step 1:
$\overrightarrow r.(2\hat i-\hat j+2\hat k)=12$
Equation of the given plane is $2x-y+2z=12$--------(1)
Direction ratios of the normal to the plane are $(2,-1,2)$
The direction cosines are
$\large\frac{2}{\sqrt{2^2+(-1)^2+(2)^2}},\large\frac{-1}{\sqrt{2^2+(-1)^2+(2)^2}},\large\frac{2}{\sqrt{2^2+(-1)^2+(2)^2}}$
$\Rightarrow \large\frac{2}{3},\frac{-1}{3},\frac{2}{3}$
Step 2:
Divide the equation (1) by 3 we get,
$\large\frac{2}{3}$$x-\large\frac{1}{3}$$y+\large\frac{2}{3}$$z=\large\frac{12}{3}$
which is of the form $lx+my+nz=d$
$d=4$
Hence the distance of the plane from the origin is 4units.