3. [0/1 Points] DETAILS PREVIOUS ANSWERS SCALCET8 12.5.040. MY NOTES PRACTICE ANC Find an equation of the plane. The plane that passes through the line of intersection of the planes x - z = 3 and y + 4z = 3 and is perpendicular to the plane x + y - 2z = 68. [-/1 Points] DETAILS SCALCET8 12.5.509.XP. MY NOTES P Find an equation of the plane. The plane that passes through the point (-2, 3, 3) and contains the line of intersection of the planes x + y - z = 3 and 3x - y + 4z = 5Sal. (a) equation of Name that pariers through the interaction of two planes P3 = P2 + + PI * n - 2 - 3 ) + d (4+ 42- 3 ) = 0 J X + yd + ( 4 z - z) x -3-3x = 0 - X + y x * + 32 x - 3 ( 1 + 2 ) = 0 new, Jon above plane to be perpendicular to the given plane x + y- 22-6= 0. Dat product of their normal vector should he ze50. - ara, + bib , + acz = O *1 x 1 + X x1 + 3x (- 2) -8(x ) = 0 1 + x - 6 * = 0 4 - Sd = 0 P3 = * + 32 - 3 ( 1 + 1 ] = 0 5nty + 32 - 3 (6) = 0 521 + 4 +32-18-0(8 ) now, Plane Pz that passer through the line of interaction of the planes P, = Xty - 2 = 3 and P, = 3x -4+ 42 =5 P3 = Pq + > Py = ( * + y - z - 3 ) + 2 ( 3 x - 4 + 4 2 - 5 ) = 0 = ) x ( 1 + 3 7 ) + y (1 - *) - z(1-4x ) - 3-SA= O P3 passers through paint ( - 2, 3, 3 ) P3 = [- 2 ) (1+ 3x ) + 3(12 x ) - 3 ( 1- 4 x ) - 3 - SA = O = - 2- 6*+ 3- 3x- 3+ 12x - 3 -5x =0 =) - 5-20 = 0 A = now P3 = X ( 1 + 3 1 - [ ] ) + 4 ( 1 + 5) - 2 (1- 4x (-{) ) - 3 - 5 x (-{ ) = X1 - 15) ! ) + y ( 7 ) - 2 ( 1 + 10 ) - 3+25 = ) - 13X + yx7- 112+ 19 Z S ) 13x - Fy+ 227-19 = 0