Let the equation of a plane be ax + by + cz + d = 0

When P and Q are on the same side of the plane

⇒ \(\frac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{a{{x}_{2}}+b{{y}_{2}}+c{{z}_{2}}+d}<0\)

⇒ \(\frac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{a{{x}_{2}}+b{{y}_{2}}+c{{z}_{2}}+d}>0\)

⇒ ax_{1} + by_{1} + cz_{1 }+ d and ax_{2} +by_{2} + cz_{2} + d are of the same sign.

When P and Q are on the opposite side of the plane

⇒ \(\frac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{a{{x}_{2}}+b{{y}_{2}}+c{{z}_{2}}+d}<0\)

⇒ ax_{1} + by_{1} +cz_{1} + d and ax_{2} + by_{2} + cz_{2}+ d are of opposite sign.

**Perpendicular Distance of a point from a plane: **The length of the perpendicular from the point P (x_{1}, y_{1}, z_{1}) on the plane ax + by + cz + d = 0 is \(\frac{\left| a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}\).

**Distance between Parallel Planes: **

Let ax_{1} + by_{1} + cz_{1} + d = 0 … (i)

And, ax_{2} + by_{2} + cz_{2} + d = 0 … (ii)

Be two parallel planes

Then, the distance between them \(\frac{\left| {{d}_{1}}-{{d}_{2}} \right|}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}\).

**Image of a Point in a Plane: **Let P an Q be two points and let π be a plane such that

i) Line PQ is perpendicular to the plane π, and

ii) Mid-point of PQ lies on the plane π. Then either of the point is the image of the other in the plane π.In order to find the image of a point in a given plane. We may use the following algorithm

**Algorithm:**

**Step I:** Write the equations of the line passing through P and normal – to the given plane as \(\frac{x-{{x}_{1}}}{a}=\frac{y-{{y}_{1}}}{b}=\frac{z-{{z}_{1}}}{c}\).

**Step II:** Write the coordinates of image Q as (x_{1} + ar, y_{1} + br, z_{1} + cr).

**Step III:** Find the coordinates of the mid-point R of PQ.

**Step IV:** Obtain the value of r by putting the coordinates of R in the equation of the plane.

**Step V:** Put the value of r in the coordinates of Q.

**Example:** Find the image of the point (3, -2, 1) in the plane 3x – y + 4z = 2.

**Solution: **Let Q be the image of the point P (3, -2, 1) in the plane 3x – y + 4z = 2. Then, PQ is normal to the plane.

Therefore, direction ratios of PQ are 3, -1, 4. Since PQ passes through P (3, -2, 1) and has direction ratios 3, -1, 4.Therefore, its equations is \(\frac{x-3}{3}=\frac{y+2}{-1}=\frac{z-1}{4}=r\).