$X$ intercept of the plane containing the line of intersection of the planes $x-2y+z+2=0$ and $3x-y-z+1=0$ and also passing through $(1,1,1)$ is

Equation of plane passing through line of intersection of given planes is given by,

$\left(x-2y+z+2\right)+\alpha \left(3x-y-z+1\right)=0$

Since, the plane passes through the point $\left(1, 1, 1\right)$ therefore, this point will satisfy equation of plane.

$\Rightarrow \left(1-2+1+2\right)+\alpha \left(3-1-1+1\right)=0$

$\Rightarrow \alpha =-1$

Then, equation of planes becomes $\left(x-2y+z+2\right)-1\left(3x-y-z+1\right)=0$

$\Rightarrow -2x-y+2z+1=0$

Now, $X$ intercept of this plane can be calculated by putting $y=z=0$, we get

$-2x+1=0$

$\Rightarrow x=\frac{1}{2}$ is the $X$ intercept.