Q. 145.0( 1 Vote )

Find the equation

Answer :

Given – A plane passes through (2, - 1, 5), perpendicular to the plane x + 2y - 3z = 7 and parallel to the line


To find – The equation of the plane


Let the equation of the required plane be ax + by + cz + d = 0……(a)


The plane passes through (2, - 1, 5)


So, 2a - b + 5c + d = 0…………………(i)


The direction ratios of the normal of the plane is given by (a, b, c)


Now, this plane is perpendicular to the plane x + 2y - 3z = 7 having direction ratios (1, 2, - 3)


So, a + 2b - 3c = 0……………(ii)


This plane is also parallel to the line having direction ratios (3, - 1, 1)


So, the direction of the normal of the required plane is also at right angles to the given line.


So, 3a - b + c = 0………………(iii)


Solving equations (ii) and (iii),


[α → arbitrary constant]





Putting these values in equation (i) we get,


2Х(- α) - (- 10α) + 5(- 7α) + d = 0 i.e. d = 27α


Substituting all the values of a, b, c and d in equation (a) we get,



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