Your equations for (a) are x-y+2z=7, 2x+y-z=3, and x+y+z=9. I think a first obvious thing to do is to add the second and third equations, eliminating z: (2x+y-z)+(x+y+z)= 3x+ 2y= 3+ 9= 12.
Multiplying the second equation by 2 gives 4x+2y-2z= 6 and adding that to the first equation also eliminates z: (x-y+2z)+ (4x+2y-2z)= 5x+y= 6+7= 13.
Now multiply that by 2 to get 10x+ 2y= 26 and subtract 3x+2y= 12.
(10x+2y)-(3x+2y)= 7x= 26- 12= 14. Since 7x= 14, x= 2. Then 3x+2y= 3(2)+ 2y= 6+ 2y= 12 so 2y= 6 and y= 3. Finally, x+y+z= 2+3+z=9 so z= 4. The point of intersection of the three planes is (2, 3, 4).
Well, you could try to solve for the intersection and see what happens!----------
however there is a part c) and this one worked out (i think?):
so this is my answer: not sure if it correct
pi1//pi2 and no common intersections
If you subtract the third equation from the second you eliminate x: (2x-3y+z)- (2x-18y-8z)= 15y+ 9z= 0. If you multiply the second equation by 2, getting 4x-6y+z= 0 and subtract the first, eliminating x again, (4x-6y+2z)-(4x-6y+2z)= 0x+0y+0z= 0= 10- 0. Because that equation reduces to "10= 0" which is false, no matter what x, y, and z are, there is no point that satisifies. Yes, you are correct. pi1 and pi2 are parallel and have no common point.