The equation of a line $4x-4y-z+11=0=x+2y-z-1$ can be put as: 

The given equation are

$4x-4y-z+11=0$       ….(i)

$x+2y-z+1=0$           …. (ii)

The DR's of normals to the planes (i) and (ii) are $4,-4,-1$and $1,2,-1$ respectively.

Let DR's ofline of intersection of plane be $\mathrm{l}.\mathrm{m}.\mathrm{n}$ As the line of intersection of the planes is perpendicular to 

the normals of the both planes, we get $4l-4m-n=0$ and $l+2m-n=0$

By cross multiplication $\frac{l}{6}=\frac{m}{3}=\frac{n}{12}$

or $\frac{l}{2}=\frac{m}{1}=\frac{n}{4}$

If $\mathrm{x} = 0$, equation (i) and (ii) becomes $-4y-z+11=0$, $2y-z-1=0$

Solving, we get $y=2,z=3$

$\therefore$Equation o me is $\frac{x}{2}=\frac{y-2}{1}=\frac{z-3}{4}$

Also $x=4,y=4,z=11$ satisfies equation (i) and (ii) Hence, (b) is also the correct option.