Question

If the point $ (2,\alpha ,\beta ) $ lies on the plane which passes through the points (3,4,2) and (7,0,6) and is perpendicular to the plane $ 2x - 5y = 15 $ , then $ 2\alpha - 3\beta $ is equal to
A 5
B 17
C 12
D 7

Answer 1

Hint: In this question we have been given the points (3,4,2) and (7,0,6). So, we will firstly write the equation of plane, $ a(x - {x_1}) + b(y - {y_1}) + c(z - {z_1}) = 0 $ for these two points, then we have been given that they are perpendicular to the plane $ 2x - 5y = 15 $ , so from this we will find the values of a, b and c. then we will find the value of $ 2\alpha - 3\beta $ using these points.

Complete step-by-step answer:
We have been provided that the point $ (2,\alpha ,\beta ) $ lies on the plane which passes through the points (3,4,2) and (7,0,6).
So, we will let the equation of the plane $ a(x - {x_1}) + b(y - {y_1}) + c(z - {z_1}) = 0 $ passing through (3,4,2).
 $ a(x - 3) + b(y - 4) + c(z - 2) = 0 - > (1) $ ,
We have been given that this pane also passes from the points (7,0,6).
So, we can substitute it for (x, y, z) coordinates in the above equation and we get, $ a(7 - 3) + b(0 - 4) + c(6 - 2) = 0 $ ,
Now simplifying the equation, we get, $ a - b + c = 0 - > (2) $ .
We have been given that eq (1) is perpendicular to plane $ 2x - 5y = 15 $ ,
So, we know that if a plane is perpendicular to another plane the formula used is, $ {a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0 $
Therefore, the equation would be, $ 2a - 5b + 0c = 0 - > (3) $ ,
Now we will solve eq (2) and (3),
 $ a - b + c = 0 $ and $ 2a - 5b + 0c = 0 $ ,
 $ \dfrac{a}{5} = \dfrac{b}{2} = \dfrac{c}{{ - 3}} = \lambda $
So, the values come out to be, $ a = 5\lambda ,b = 2\lambda ,c = - 3\lambda $ ,
Putting these values in eq (1), we get, $ 5\lambda (x - 3) + 2\lambda (y - 4) - 3\lambda (z - 2) = 0 $ ,
The equation would be, $ 5x + 2y - 3z = 17 - > (4) $ ,
This is the required equation of the plane.
Now, we have been given that point $ (2,\alpha ,\beta ) $ lies on this plane eq (4),
We will get, $ 5(2) + 2(\alpha ) - 3(\beta ) = 17 $
So, from this we can find the value of $ 2\alpha - 3\beta $ , $ 2\alpha - 3\beta = 7 $ ,

So, the correct answer is “Option d”.

Note: We can solve this question by another method as we have been given that the points (3,4,2) and (7,0,6) and is perpendicular to the plane $ 2x - 5y = 15 $ so we can find the cross product to find the equation of the plane, and after that we can simply put the points $ (2,\alpha ,\beta ) $ in that equation to get the value of $ 2\alpha - 3\beta $ .