# 9 Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1).

The equation of any plane through the intersection of the planes,

$3x-y+2z-4=0\ and\ x+y+z-2=0$

Can be written in the form of; $(3x-y+2z-4)\ +\alpha( x+y+z-2)= 0$, where $\alpha \epsilon R$

So, the plane passes through the point $(2,2,1)$, will satisfy the above equation.

$(3\times2-2+2\times1-4)+\alpha(2+2+1-2) = 0$

That implies $2+3\alpha= 0$

$\alpha = \frac{-2}{3}$

Now, substituting the value of $\alpha$ in the equation above we get the final equation of the plane;

$(3x-y+2z-4)\ +\alpha( x+y+z-2)= 0$

$(3x-y+2z-4)\ +\frac{-2}{3}( x+y+z-2)= 0$

$\Rightarrow 9x-3y+6z-12\ -2 x-2y-2z+4= 0$

$\Rightarrow 7x-5y+4z-8= 0$ is the required equation of the plane.

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