Question

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).

Answer 1

I hope it will be answer of your question

Answer 2

Answer:

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

Step-by-step explanation:

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).

$\text{Given equations are}$

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

$\text{The equation of the plane of the plane passing through }$$\text{the point of intersection of the given two planes is}$

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

$\text{It passes through (2,2,1)}$

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

$(6-2+2-4)+\lambda(2+2+1-2)=0$

$(2)+\lambda(3)=0$

$\implies\bf\lambda=\frac{-2}{3}$

$\text{The required plane is }$

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

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

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

$\implies\boxed{\bf\:7x-5y+4z-8=0}$