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Q.
Equation of plane passing through the points $ (2,\,\,2,\,\,1) $ , $ (9,\,\,3,\,\,6) $ and perpendicular to the plane $ 2x+6y+6z-1=0 $ , is
Jharkhand CECEJharkhand CECE 2015
Solution:
Equation of a plane passing through $ (2,\,\,2,\,\,1) $ is $ a(x-2)+b(y-2)+c(z-1)=0 $ ... (i)
This passes through $ (9,\,\,3,\,\,6) $ and is perpendicular to $ 2x+6y+6z-1=0 $
$ \therefore $ $ 7a+b+5c=0 $ and $ 2a+6b+6c=0 $
On solving above equations by cross-multiplication, we get
$ \Rightarrow $ $ \frac{a}{6-30}=\frac{-b}{42-10}=\frac{c}{42-2} $
$ \Rightarrow $ $ \frac{a}{-24}=\frac{b}{-32}=\frac{c}{40} $
$ \Rightarrow $ $ \frac{a}{-3}=\frac{b}{-4}=\frac{c}{5} $
On substituting the values of
$ a,\,\,b $ and $ c $ in Eq. (i),
we get $ -3(x-2)-4(y-2)+5(z-1)=0 $ or $ 3x+4y-5z-9=0 $
as the required plane.