# Prove that the image of the point (3,-2,1) in the plane $$3x-y+4z=2$$ lies on the plane $$x+y+z+4=0$$.

Toolbox:
• Mid point of AB where $A(x_1,y_1,z_1)\:and\:B(x_2,y_2,z_2)$ is given by $\bigg(\large\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\bigg)$
Step 1:
Equation of the given plane is $3x-y+4z=2$----(1)
Let $P(3,-2,1)$ have image as $P'(x_1,y_1,z_1)$
Thus PP' is $\perp$ to the plane (1)
Let $M$ be the mid point of PP' and lie on th plane (1)
Equation of the line PP' is
$\large\frac{x-3}{3}=\frac{y+2}{-1}=\frac{z-1}{4}$$=k(say) Any point P' on it is (3k+3,-k-2,4k+1) Since this point lies on the plane 3(3k+3)-(-k-2)+4(4k+1)=2 \therefore k=-\large\frac{1}{2} Step 2: Now substituting the value of k we get the coordinate of M as (3(-\large\frac{1}{2})$$+3,\large\frac{1}{2}$$-2.4(-\large\frac{1}{2})$$+1)$
$\Rightarrow (\large\frac{3}{2},-\frac{3}{2}$$,-1) Since M is the mid point of PP' \large\frac{3+x_1}{2}=\frac{3}{2} \Rightarrow x_1=0 \large\frac{-2+y_1}{2}=\frac{-3}{2} \Rightarrow y_1=-1 \large\frac{1+z_1}{2}=$$-1$
$z_1=-3$
Step 3:
Hence the coordinates of M are (0,-1,-3)
Now substituting for $x,y$ and $z$ in the equation
$x+y+z+4=0$
$\Rightarrow 0-1-3+4=0$
Hence the image of the point $(3,-2,1)$ in the plane $3x-y+4z=2$ lies on the plane.
$x+y+z+4=0$