The image of the point $(3,-2,1)$ in the plane $3 x-y+4 z=2$ is:

$(0,-1,-3)$

The correct answer is Option (2) → $(0,-1,-3)$

The foot of a point on the plane is,

$Q(x',y',z')=(x-λA,y-λB,z-λC)$

$P(x,y,z)=(3,-2,1)$

$Ax+By+Cz=(A=3,B=-1,C=4,D=2)$

λ → Scalar

The perpendicular distance 'd' from the point $(x,y,z)$,

$d=\frac{|Ax+By+Cz-D|}{\sqrt{A^2+B^2+C^2}}$

$=\frac{|3(3)-(-2)+4(1)-2|}{\sqrt{3^2+(-1)^2+4^2}}=\frac{13}{\sqrt{26}}=\frac{\sqrt{26}}{2}$

and,

$λ=\frac{Ax+By+Cz-D}{A^2+B^2+C^2}=\frac{13}{26}=\frac{1}{2}$

$Q(x',y',z')=\left(3-\frac{1}{2}×3,-2-\frac{1}{2}×(-1),1-\frac{1}{2}×4\right)$

$=\left(3-\frac{3}{2},-2+\frac{1}{2},1-2\right)$

$=\left(\frac{3}{2},-\frac{3}{2},-1\right)$

Image = $2Q-P$

$=2\left(\frac{3}{2},-\frac{3}{2},-1\right)-(3,-2,1)$

$=(0,-1,-3)$