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**Hint:**Here, we will use the formula of the equation of the plane passing through the intersection of two planes. Then we will substitute the value of a point lying on this plane using the fact that it is perpendicular to the\[xy\] plane. From this we will get the value of \[\lambda \], and substituting its value, we will get the required equation of the plane.

**Formula Used:**

Equation of passing through intersection of two plane \[ = \left( {\,{A_1}x + {B_1}y + {C_1}z - {d_1}} \right) + \lambda \left( {{A_2}x + {B_2}y + {C_2}z - {d_2}} \right)\]

**Complete step-by-step answer:**We know that, equation of passing through intersection of two planes which are in the form of \[{A_1}x + {B_1}y + {C_1}z = {d_1}\] and \[{A_2}x + {B_2}y + {C_2}z = {d_2}\] is given as:

\[\left( {\,{A_1}x + {B_1}y + {C_1}z - {d_1}} \right) + \lambda \left( {{A_2}x + {B_2}y + {C_2}z - {d_2}} \right)\]…………………………………\[\left( 1 \right)\]

Comparing the general equation of two planes by the given planes, we get,

\[{A_1}x + {B_1}y + {C_1}z - {d_1} = ax + by + cz + d\]

and

\[{A_2}x + {B_2}y + {C_2}z - {d_2} = \alpha x + \beta y + \gamma z + e\]

Hence, from equation \[\left( 1 \right)\], we get

Equation of the planes passing through intersection of given planes \[ = \left( {ax + by + cz + d} \right) + \lambda \left( {\alpha x + \beta y + \gamma z + e} \right)\]

Opening the brackets and taking the like variables common, we get,

\[ \Rightarrow \] Equation of the plane \[ = \left( {a + \lambda \alpha } \right)x + \left( {b + \lambda \beta } \right)y + \left( {c + \lambda \gamma } \right)z + d + \lambda e = 0\]……………….\[\left( 2 \right)\]

Also, according to the question, this plane is also perpendicular to \[xy\] plane.

Hence, if a plane is perpendicular to \[xy\] plane then it means that it lies on \[z\] plane.

Then, its coordinates are of the form \[\left( {0,0,1} \right)\], where \[x\] and \[y\] coordinates are 0 and \[z\] coordinate is 1.

Hence, substituting \[x = 0\], \[y = 0\] and \[z = 1\] in \[\left( {a + \lambda \alpha } \right)x + \left( {b + \lambda \beta } \right)y + \left( {c + \lambda \gamma } \right)z\], we get

\[\left( {a + \lambda \alpha } \right)\left( 0 \right) + \left( {b + \lambda \beta } \right)\left( 0 \right) + \left( {c + \lambda \gamma } \right)\left( 1 \right) = 0\]

\[ \Rightarrow 0 + 0 + \left( {c + \lambda \gamma } \right) = 0\]

Now, rewriting this as:

\[ \Rightarrow \lambda = \dfrac{{ - c}}{\gamma }\]

Now, substituting \[\lambda = \dfrac{{ - c}}{\gamma }\] in equation \[\left( 2 \right)\], we get

\[\left( {a + \left( {\dfrac{{ - c}}{\gamma }} \right)\alpha } \right)x + \left( {b + \left( {\dfrac{{ - c}}{\gamma }} \right)\beta } \right)y + \left( {c + \left( {\dfrac{{ - c}}{\gamma }} \right)\gamma } \right)z + d + \left( {\dfrac{{ - c}}{\gamma }} \right)e = 0\]

Taking LCM and solving further, we get

\[ \Rightarrow \left( {a\gamma - c\alpha } \right)x + \left( {b\gamma - c\beta } \right)y + \left( {c\gamma - c\gamma } \right)z + d\gamma - ce = 0\]

\[ \Rightarrow \left( {a\gamma - c\alpha } \right)x + \left( {b\gamma - c\beta } \right)y + d\gamma - ce = 0\]

Hence, the equation of the plane through the line of intersection of the planes is \[\left( {a\gamma - c\alpha } \right)x + \left( {b\gamma - c\beta } \right)y + \left( {d\gamma - ce} \right) = 0\].

**Therefore, option A is the correct answer.**

**Note:**A plane is a two dimensional flat surface in which if any two random points are chosen then, the straight line joining those points completely lie on that plane. A plane is defined by three points unless they are forming a straight line. In a scalar equation, when a plane passes through the intersection of two planes, then we combine their equations to form one single equation of the required plane.

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