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.

$x+y+z+4=0$