The direction ratio's of line of intersection of two planes :

2x + y + z + 47 = 0 and 3x - 2y - z + 41 = 0 are :

< 1, 5, - 7 >

Normal vector to plane ax + by + cz + d = 0 is represents by ai + bj + ck. 

Hence, normal to first plane is i + j + k and normal to second plane is given by 3i - 2j - k. 

The line of intersection plane ⊥ to both these normal vectors. 

It's found by cross product of these normal vectors. 

$(2i + j + k)×(3i - 2j - k)=i−2j+k=\begin{bmatrix}\hat i & \hat j & \hat k\\2&1&1\\3&-2&-1\end{bmatrix}$

$\begin{bmatrix}\hat i & \hat j & \hat k\\2&1&1\\3&-2&-1\end{bmatrix}=\hat i(-1+2)-\hat j(-2-3) + \hat k(-4-3)$

$\hat i - \hat j(-5)+\hat k(-7) ⇒\hat i + 5\hat j - 7\hat k$

Direction ratio's are <1, 5, -7>