Different Forms of Equation of a Plane

Let R, S and T be three non collinear points on the plane with position vectors `vec a` , `vec b` and `vec c` respectively in following fig.

The vectors `vec (RS)` and `vec (RT)` are in the given plane. Therefore , the vector `vec (RS) xx vec (RT)` is perpendicular to the plane containing points R,S and T. Let `vec r`be the position vector of any point P in the plane.  Therefore, the equation of the plane passing through R and  perpendicular to the vector  `vec (RS) xx vec (RT)` is 
`(vec r - vec a) . (vec (RS) xx vec (RT)) = 0`
or `(vec r - vec a) . [(vec b - vec a) xx (vec c - vec a)] = 0`  ...(1)
If the three points were on the same line, then there will be many planes that will contain them Fig.

for example , These planes will resemble the pages of a book where the line containing the points R, S and T are members in the binding of the book.

Cartesian form:
Let `(x_1,y_1,z_1) , (x_2 , y_2 , z_2)` and `(x_3 , y_3 , z_3)`  be the coordinates of the points R, S and T respectively.  Let (x, y, z) be the coordinates of any point P on the plane with position vector `vec r`. Then  
`vec (RP) = (x - x_1) hat i + ( y - y_1) hat j + (z - z_1) hat k`
`vec (RS) = (x_2 - x_1) hat i + ( y_2 - y_1) hat j + (z_2 - z_1) hat k`
`vec (RT) = (x_3 - x_1) hat i + ( y_3 - y_1) hat j + (z_3 - z_1) hat k`
Substituting these values in equation (1) of the vector form and expressing it in the form of a determinant, we have 
`|(x - x_1 , y - y_1 , z - z_1),(x_2 - x_1 , y_2 - y_1, z_2 -z_1) ,(x_3 - x_1, y_3 - y_1, z_3 - z_1) | = 0`
which is the equation of the plane in Cartesian form passing through three non collinear points `(x_1, y_1, z_1), (x_2, y_2, z_2)` and `(x_3, y_3, z_3).`

Video link : https://youtu.be/PCyo3E5kOcw

Equation of a Plane in Intercept form:

\[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\]

Distance of the Plane from Origin is

\[d=\frac{1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}}\]

Consider a plane whose perpendicular distance from  the origin is d (d ≠ 0). in following fig. 

If `vec (ON)` is the normal from the origin to the plane, and `hat n ` is the unit normal vector along `vec (ON)`. Then `vec (ON)` = d . `hat n` . Let P be any point on the plane. Therefore , `vec (NP)` is perpendicular to `vec (ON)`.
Therefore, `vec (NP) . vec (ON) = 0`  ...(1)
Let `vec r` be the position vector of the point P, then `vec (NP) = vec r - d . hat n`     (as `vec (ON) + vec (NP) = vec (OP)`)
Therefore, (1) becomes 
`(vec r - d . hat n) . d hat n = 0`
or `(vec r - d.hat n). hat n =0`  (d ≠ 0)
or `vec r . hat n - d  hat n . hat n = 0`
i.e., `vec r .  hat n = d`        (as `hat n . hat n = 1`)          ...(2)
This is the vector form of the equation of the plane. 

Cartesian form
Equation (2) gives the vector equation of a plane, where `hat n` is the unit vector normal to the plane. Let P(x, y, z) be any point on the plane.  Then 
`vec (OP) = vec r = x hat i + y hat j + z hat k`
Let l, m, n be the direction cosines of `hat n` . Then
`hat n = l hat i + m hat j + n hat k ` 
Therefore, (2) gives
`(x hat i + y hat j + z hat k) . (l hat i + m hat j + n hat k) = d`
i.e. lx + my + nz = d                             ... (3) 

There can be many planes that are perpendicular to the given vector, but through a given point `P(x_1, y_1, z_1)`, only one such plane exists in following fig. 

Let a plane pass  through a point A with position vector `vec a` and perpendicular to the vector
`vec N `
Let `vec r` be the position vector of any point P(x,y,z) in the plane. Fig.

Then the point P lies in the plane if and only if `vec (AP)` is perpendicular to `vec N` . i.e., `vec (AP) . vec N = 0` . But `vec (AP) = vec r - vec a.` Therefore `(vec r - vec a) . vec N = 0`  ...(1)

Cartesian form: 
Let the given point A be `(x_1,y_1,z_1)` ,P be (x , y, z) and direction ratios of `vec N` are A ,B and C . Then,
`vec a = x_1 hat i + y_1 hat j + z_1 hat k ,   hat r = x hat i + y hat j +z hat k`  and  `vec N = A hat i + B hat j + C hat k`
Now `(vec r - vec a) . vec N = 0`
So `[(x - x_1) hat i + (y - y_1) hat j + (z - z_1) hat k] . (A hat i + B hat j +C hat k) = 0`
i.e. `A (x - x_1) + B (y - y_1) + C (z - z_1) = 0`

Video link : https://youtu.be/q6ASCg_D_gY

Let `π_1` and `π_2` be two planes with equations `vec r  . hat n _1 = d_1` and `vec r . hat n _2 = d_2` respectively.  The position vector of any point on the line of intersection must satisfy both the equations fig.

If `vec t ` is the position vector of a point on the line , then
`vec t . hat n_1 = d_1` and `vec t . hat n _2 = d_2` 
Therefore , for all real values of  λ, we have
`vec t . (hat n _1 + lambda hat n_2) = d_1 + lambda d_2`
Since `vec t` is arbitrary, it satisfies for any point on the line.
Hence , the equation `vec r . (vec n_1 + lambda vec n_2) = d_1 + lambda d_2`   represents a plane `π_3` which is such  that if any vector ` vec r` satisfies both the equations `π_1` and `π_2`, it also satisfies the equation `π_3` i.e., any plane passing through the intersection of the planes 
`vec r . vec n_1 = d_1` and `vec r . vec n_2 = d_2`
has the equation  `vec r . (vec n_1 + lambda vec n_2) = d_1 + lambda d_2`    ...(1)

Cartesian form:
In Cartesian system, let  `vec n_1 = A_1 hat i + B_2 hat j + C_1 hat k`
`vec n_2 = A_2 hat i + B_2 hat j + C _2 hat k` 
and `vec r = x hat i + y hat j + z hat k`
Then (1) becomes 
`x (A_1 + lambda A_2) + y(B_1 + lambda B_2) + z(C_1 +lambda C_2) = d_1 + lambda d_2`
or `(A_1x +B_1y + C_1z -d_1) + lambda (A_2x + B_2y + C_2z -d_2) = 0`     ..(2)
which is the required Cartesian form of the equation of the plane passing through the intersection of the given planes for each value of λ.