Let us take z=0, then
x−y=1
3x=3
⟹x=1,y=1
Thus the point is (1,1,0)
when x=0, then
2y−z=−1
y−z=2
Multiplying by 3, we get
3y−3z=6
−y+3z=1
2y=7
⟹y=72,z=32
Thus the point is (0,72,32)
Let a be the vector from (−1,2,1) to(1,1,0)
a=⟨1−(−1),1−2,0−1⟩=⟨2,−1,−1⟩
Let b be the vector from (−1,2,1) to (0,7/2,3/2):
b=⟨0−(−1),72−2,32−1⟩=⟨1,32,12⟩
The normal to the plane is:
n=a×b=∣∣
∣
∣
∣∣ijk2−1−113212∣∣
∣
∣
∣∣=1i−2j+4k
Hence, the equation of plane is given\quad by
n∗⟨x−x0,y−y0,z−z0⟩=0
⟹⟨1,−2,4⟩∗⟨x−(−1),y−2,z−1⟩=0
⟹1(1+x)−2(y−2)+4(−1)=0
⟹x−2y+4=−1 which is the required equation of plane.