The equation of the plane through the intersection of the planes x + y + z = 1 and 2x + 3y - z + 4 = 0 and parallel to x-axis, is

$y - z + 6 = 0 $

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 + (3 λ +1) y + (1- λ) x + 4λ - 1 = 0 $ .......(i)

It is parallel to X-axis i.e., $\frac{x}{1}=\frac{y}{0}=\frac{z}{0}$

$∴ 1(2λ + 1) + 0 × (3 λ + 1) + 0 ( 1 - λ )= 0 ⇒ λ =-\frac{1}{2}$

Substituting $ λ = -\frac{1}{2}$ in (i), we get

$y - 3z + 6 = 0 $ as the equation of the required plane.