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

Hint: Assume that the coordinates of the image of the given point is (x, y, z). Apply the formula for coordinates of image of a given point \[\left( \alpha ,\beta ,\gamma \right)\] in the plane $ax + by + cz + d = 0$ given as: - \[\dfrac{\alpha -x}{a}=\dfrac{\beta -y}{b}=\dfrac{\gamma -z}{c}=2\left( \dfrac{a\alpha +b\beta +c\gamma +d}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}} \right)\]. Once, (x, y, z) are determined, substitute it in the plane $x + y + z + 4 = 0$. If the point satisfies this plane then it lies on the plane otherwise not.

Complete step-by-step solution
Here, we have been provided with coordinates of a point as (3, -2, 1) and we have to prove that its image in the plane $3x – y + 4z = 2$ will satisfy the equation of plane $x + y + z + 4 = 0$, that means the image will lie on this plane. So first, let us find the coordinates of the image of the given point.
Let us assume the coordinates of the image of the point (3, -2, 1) in the plane $3x – 4y + 4z = 2$ is given as (x, y, z). Now, we know that coordinates of image of a point \[\left( \alpha ,\beta ,\gamma \right)\] in the plane $ax + by + cz + d = 0$ is given by the formula: -
\[\Rightarrow \dfrac{\alpha -x}{a}=\dfrac{\beta -y}{b}=\dfrac{\gamma -z}{c}=2\left( \dfrac{a\alpha +b\beta +c\gamma +d}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}} \right)\]
So, considering the given point (3, -2, 1) as \[\left( \alpha ,\beta ,\gamma \right)\] and writing the equation of the plane $3x – y + 4z = 2$ in the form $ax + by + cz + d = 0$, we get,
\[\Rightarrow 3x-y+4z-2=0\]
So, we have, a = 3, b = -1, c = 4 and d = -2
Therefore, applying the formula we get,
\[\begin{align}
  & \Rightarrow \dfrac{3-x}{3}=\dfrac{-2-y}{-1}=\dfrac{1-z}{4}=2\left(\dfrac{3\times 3+\left( -1 \right)\times \left( -2 \right)+1\times 4+\left( -2 \right)}{{{3}^{2}}+{{\left( -1 \right)}^{2}}+{{4}^{2}}} \right) \\
 & \Rightarrow \dfrac{3-x}{3}=\dfrac{2+y}{1}=\dfrac{1-z}{4}=2\times \left( \dfrac{9+2+4-2}{9+1+16} \right) \\
 & \Rightarrow \dfrac{3-x}{3}=\dfrac{2+y}{1}=\dfrac{1-z}{4}=2\times \dfrac{13}{13\times 2} \\
 & \Rightarrow \dfrac{3-x}{3}=\dfrac{2+y}{1}=\dfrac{1-z}{4}=1 \\
\end{align}\]
Considering \[\dfrac{3-x}{3}=1\] and cross – multiplying, we get,
\[\begin{align}
  & \Rightarrow 3-x=3 \\
 & \Rightarrow -x=0 \\
 & \Rightarrow x=0 \\
\end{align}\]
Considering \[\dfrac{2+y}{1}=1\] and cross – multiplying, we get,
\[\begin{align}
  & \Rightarrow 2+y=1 \\
 & \Rightarrow y=1-2 \\
 & \Rightarrow y=-1 \\
\end{align}\]
Considering \[\dfrac{1-z}{4}=1\] and cross – multiplying, we get,
\[\begin{align}
  & \Rightarrow 1-z=4 \\
 & \Rightarrow z=1-4 \\
 & \Rightarrow z=-3 \\
\end{align}\]
Therefore, the coordinates of the image of the given point (3, -2, 1) is (x, y, z) = (0, -1, -3).
Now, we have to prove that this image will lie on the plane $x + y + z + 4 = 0$. If this image lies on the plane $x + y + z + 4 = 0$ then it will satisfy this plane. So, substituting (0, -1, -3) in the equation of this plane, we get,
\[\Rightarrow \]L.H.S = $x + y + z + 4 $
\[\Rightarrow \]L.H.S = $0 – 1 – 3 + 4$
\[\Rightarrow \]L.H.S = $0$
\[\Rightarrow \]L.H.S = R.H.S
Hence, the point of the image satisfies the plane so it can be concluded that the image lies on the plane $x + y + z + 4 = 0$. Hence proved.

Note: One may note that the formula for the image of a point in a plane is analogous to that of the formula for the image of a point in a line. The only difference is that a line lies on a plane, that means a plane is a collection of lines. You must remember the formula to determine the coordinates of the image of a point otherwise it will be very difficult for us to solve the question. Note that the coordinates of foot of perpendicular of a point is also given by an almost similar formula given as: - \[\dfrac{\alpha -x}{a}=\dfrac{\beta -y}{b}=\dfrac{\gamma -z}{c}=2\left( \dfrac{a\alpha +b\beta +c\gamma +d}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}} \right)\]. So, do not get confused in both the formulas.