$\begin{array}{1 1} 7x-5y+4z-8=0 \\ 11x-y+8z-16=0 \\ 7x-5y+4z+8=0 \\ 7x-5y-4z-8=0\end{array} $

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- The equation of a plane passing through the intersection of two planes is $(a_1x_1+b_1y_1+c_1z_1+d_1)+\lambda(a_2x_2+b_2y_2+c_2z_2+d_2)=0$
- Where $\lambda$ is any real number.

Step 1:

Given planes are $3x-y+2z=4$ and $x+y+z=2$

We know equation of a plane passing through the intersection of two planes is $(a_1x_1+b_1y_1+c_1z_1+d_1)+\lambda(a_2x_2+b_2y_2+c_2z_2+d_2)=0$

Where $\lambda$ is any real number.

Therefore $(3x-y+2x-4)+\lambda(x+y+z-2)=0$-------(1)

The given plane passes through the point $(2,2,1)$

Now substituting for $x,y$ and $z$ we get,

$(3\times 2-2+2\times 1-4)+\lambda(2+2+1-2)=0$

$\Rightarrow (6-2+2-4)+\lambda(3)=0$

Step 2:

On simplifying we get

$2+3\lambda=0$

Therefore $\lambda=\large\frac{-2}{3}$

Now substituting the value of $\lambda$ in equ(1)

We get

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

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

On simplifying we get

$7x-5y+4z-8=0$

This is the required equation of the plane.

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