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# Find the equation of the plane passing through the points ( 1, 2, 3 ) and (0, -1, 0 ) and parrallel to the line, $\large\frac{x-1}{2} = \frac{y+2}{3} = \frac{z}{-3}$.

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A)
Equation of the plane passing through  point (1,2,3) is

a(x-1)+ b(y-2)+c(z-3)=0 ..........(1)

Equation of the plane passes through (0,-1,0)

Hence putting in equation in (1)

-a-3b-3c=0.............(2)

The equation of the plane parrallal to the given line

(x-1)/2=(y+2)/3=(z-0)/-3

Normal to the plane is parpendicular to the given line

So 2a+3b-3c=0.........(3)

a=6k, b=-3k, c=k

Putting the value in equation (1)

6k(x-1)+-3k(y-2)+k(z-3)=0

=> 6x-6+3y-6+z-3=0

=> 6x+3y+z=9