Question

The equation of the plane passing through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z = 1 is.
A. 2x - 4y + 5z - 9 = 0
B. 3x + 4y - z - 5 = 0
C. 3x + 4y - 5z - 9 = 0
D. x + 4y -9z - 3 = 0

Answer 1

Hint: In this question it is given that we have to find the equation of the plane which passes through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z = 1. So to find the solution we need to know that if any plane passing through the point (p, q, r) then the equation can be written as,
$a\left( x-p\right) +b\left( x-q\right) +c\left( z-r\right) =0$....(1)
And if two plane $a_{1}x+b_{1}y+c_{1}z=0$ and $a_{2}x+b_{2}y+c_{2}z=d$ are perpendicular then,
$a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0$.....(2)

Complete step-by-step answer:
Since a plane is passing through the point (2, 2, 1) then the equation of the plane will be,
$a\left( x-2\right) +b\left( y-2\right) +c\left( z-1\right) =0$.....(3)
Where, a, b , c are the constant coefficient and we have to find the values of a, b and c in order to find the solution.
Also the equation (3) passes through the point (9, 3, 6), then the point must satisfy the equation,
$a\left( 9-2\right) +b\left( 3-2\right) +c\left( 6-1\right) =0$
$\Rightarrow 7a+b+5c=0$........(4)
Now since, equation (3) is perpendicular to the $2x + 6y + 6z = 1$
So by the formula (2) we can write,
$a\times 2+b\times 6+c\times 6=0$
$\Rightarrow 2a+6b+6c=0$
$\Rightarrow a+3b+3c=0$........(5)
Now from (4) and (5) we can write,
$\dfrac{a}{1\times 3-3\times 5} =\dfrac{b}{1\times 5-3\times 7} =\dfrac{c}{3\times 7-1\times 1}$ [by method of cross-multiplying]
$\Rightarrow \dfrac{a}{3-15} =\dfrac{b}{5-21} =\dfrac{c}{21-1}$
$\Rightarrow \dfrac{a}{-12} =\dfrac{b}{-16} =\dfrac{c}{20}$
$\Rightarrow \dfrac{a}{-3} =\dfrac{b}{-4} =\dfrac{c}{5}$ [ multiplying by 4]
Now, let,
$\dfrac{a}{-3} =\dfrac{b}{-4} =\dfrac{c}{5} =k$
$\therefore a=-3k,\ y=-4k,\ c=5k$
Now putting the value of a, b and c in equation (3) we get,
$-3k\left( x-2\right) +\left( -4k\right) \left( y-2\right) +5k\left( z-1\right) =0$
$\Rightarrow -3\left( x-2\right) -4\left( y-2\right) +5\left( z-1\right) =0$[dividing both side by k]
$\Rightarrow -3x+6-4y+8+5z-5=0$
$\Rightarrow -3x-4y+5z-5+6+8=0$
$\Rightarrow -3x-4y+5z+9=0$
$\Rightarrow 3x+4y-5z-9=0$ [ multiplying both side by (-1)]
Therefore the equation of place is $3x+4y-5z-9=0$
Hence the correct option is option C.

Note: While solving two linear equations we have used the multiplication method which states that, if $a_{1}x+b_{1}y+c_{1}z=0$ and $a_{2}x+b_{2}y+c_{2}z=0$ be two linear equations then in order to solve the equations, we can write,
$\dfrac{x}{\left( b_{1}c_{2}-b_{2}c_{1}\right) } =\dfrac{y}{\left( c_{1}a_{2}-c_{2}a_{1}\right) } =\dfrac{z}{\left( a_{1}b_{2}-a_{2}b_{1}\right) }$ .