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

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 $2\alpha - 3\beta$ .