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

The given equation are 4x – 4y –z + 11 = 0 .....…(i)
x + 2y – z – 1 = 0...….(ii) The D.r’s of normals to the planes (i) and (ii) are 4, –4, –1 and 1, 2, –1 respectively. Let Dr’s of line of intersection of plane
be l, m, n. As the line of intersection of the planes isperpendicular to the normals of the both planes, we get 4l – 4m – n = 0 and l + 2m – n = 0 By cross multiplication

$\frac{1}{6}=\frac{m}{3}=\frac{n}{12}\text{ or }\frac{1}{2}=\frac{m}{1}=\frac{n}{4}$

If x=0, Eqs. (i) and (ii) becomes –4y–z+11= 0

$\Rightarrow 2y-z-1=0\text{, Solving, we get }y=2,z=3$

$\Rightarrow \text{ Equation of line is }\frac{x}{2}=\frac{y-2}{1}=\frac{z-3}{4}$

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