The equation of plane passing through a point \[A\left( {2, - 1,3} \right)\] and parallel to the vectors \[a = \left( {3,0, - 1} \right)\;\] and \[b = \left( { - 3,2,2} \right)\;\] is [Orissa JEE \[2005\]]
A) \[\;\;2x - 3y + 6z - 25 = 0\]
B) \[\;\;2x - 3y + 6z + 25 = 0\]
C) \[\;3x - 2y + 6z - 25 = 0\]
D) \[\;3x - 2y + 6z + 25 = 0\]

Hint: in this question we have to equation of the plane for this apply the formula of general equation of plane. By using given condition find the direction ratio of vector normal to the plane. Substitute value of direction ratio in general equation of plane.

Formula Used: General equation of plane is given by
\[A(x - {x_1}) + B(y - {y_1}) + C(z - {z_1}) = 0\]
Where
A, B, C are direction ratio of vector which is normal to the plane
\[({x_1},{y_1}{z_1})\]Coordinate through which plane passing.

Complete step by step solution: Given: Coordinates through which plane passes and points to which plane is parallel.
Now required equation of plane is given by
\[A(x - {x_1}) + B(y - {y_1}) + C(z - {z_1}) = 0\]
\[A(x - 2) + B(y + 1) + C(z - 3) = 0\]
Plane is parallel to \[a = \left( {3,0, - 1} \right)\;\]and \[b = \left( { - 3,2,2} \right)\;\]
Normal to the plane is given by
\[Ai + Bj + Ck = 0\]
Where
A, B, C are direction ratio of vector which is normal to the plane
It is given in the question that plane is parallel to \[a = \left( {3,0, - 1} \right)\;\]and \[b = \left( { - 3,2,2} \right)\;\]
Therefore,
Dot product of normal and a, dot product of normal and b is zero
\[(Ai + Bj + Ck).(3i + 0j - 1k) = 0\]
\[3A - C = 0\]…………………… (i)
\[(Ai + Bj + Ck).( - 3i + 2j + 2k) = 0\]
\[ - 3A + 2B + 2C = 0\]…………… (ii)
From equation (i) and (ii) we get direction ratio of vector normal to the plane
\[A = 2,B = - 3,C = 6\]
Put these values in \[A(x - 2) + B(y + 1) + C(z - 3) = 0\]
Now required equation of plane is given by
\[2(x - 2) - 3(y + 1) + 6(z - 3) = 0\]
\[2x - 4 - 3y - 3 + 6z - 18 = 0\]
\[2x - 3y + 6z - 25 = 0\]

Option ‘A’ is correct

Note: Here we must remember that if plane is parallel to given point then normal to the plane and normal to that point is same. Dot product of perpendicular vector is always zero.