The equation of the plane through the intersection of the planes x+y+z=1 and 2x+3y−z+4=0 is (x+y+z−1)+λ(2x+3y−z+4)=0 or (2λ+1)x+(3λ+1)y+(1−λ)z+4λ−1=0 .....(i) It is parallel to X-axis. i.e.
x
1
=
y
0
=
z
0
∴ 1(2λ+1)+0×(3λ+1)+0(1−λ)=0 ⇒ λ=−
1
2
On substituting λ=−
1
2
in Eq. (i), we get (2x−
1
2
+1)x+(3×
−1
2
+1)y+(1+
1
2
)z+4×(−
1
2
)−1=0 ⇒
−1
2
y+
3
2
z−3=0 ⇒ y−3z+6=0 which is the equation of the required plane.