A straight line passes through the point (2, –1, –1). It is parallel to the plane 4x + y + z + 2 = 0 and is perpendicular to the line $\frac{x}{1}=\frac{y}{-2}=\frac{z-5}{1}$. The equations of the straight line are

$\frac{x-2}{-1}=\frac{y+1}{1}=\frac{z+1}{3}$

Let the direction cosines of the straight line be l, m, n ⇒ 4l + m + n = 0,

$l – 2m + n = 0 ⇒\frac{l}{3}=\frac{m}{-3}=\frac{n}{-9}⇒\frac{l}{-1}=\frac{m}{+1}=\frac{n}{3}$.

Equations of the straight line are $\frac{x-2}{-1}=\frac{y+1}{1}=\frac{z+1}{3}$.

Hence (C) is the correct choice.