Equation of the line passing through $(1, 1, 1)$ and parallel to the plane $2x + 3y + z + 5 = 0$ is

A line parallel to the plane must have a direction vector perpendicular to the plane's normal vector. For the plane $2x + 3y + z + 5 = 0$, a normal vector is $\left(2,3,1\right)$. So the direction vector $\left(a,b,c\right)$ of the line must satisfy $$2a+3b+c=0.$$ Now check the direction ratios in the options. For option $\mathrm{A}$, the direction vector is $\left(-1,1,-1\right)$. $$2\left(-1\right)+3\left(1\right)+\left(-1\right)=0.$$ So this direction vector is perpendicular to the normal, hence the line is parallel to the plane. It also passes through $\left(1,1,1\right)$, so its symmetric equation is $$\frac{x-1}{-1}=\frac{y-1}{1}=\frac{z-1}{-1}.$$ Re-checking the options, only option $\mathrm{A}$ satisfies the required condition.