Are there scalars which makes $u_1$ to $u_3$ linearly independent? There are 4 vectors:
$v_1 = (1,4,2,-3) , v_2 = (-2,4,-4,27) , v_3 = (1,8,2,4) , v_4 = (2,-12,4,-41)$

Are there any scalars $a_i, b_i, c_i ( 1 \le i \le 4)$ that the vectors 

$u_1 = \sum\limits_{i=1}^4 a_iv_i$  

$u_2 = \sum\limits_{i=1}^4 b_iv_i$ 

$u_3 = \sum\limits_{i=1}^4 c_iv_i$

be independent?

Well, I put $v_1,v_2,v_3,v_4$ on a matrix and got 3 vectors which are linearly independent.

Which means that if I want $u_1,u_2,u_3$ to be linearly independent I will need to put for every $u_i$ vector two scalars to be zero.

for example:

$u_1 = a_1v_1$

$u_2 = b_2v_2$

$u_3 = c_3v_3$

and like that I will have about 6 options I guess to make linearly independency. What do you think? Did I solve this all wrong and there's a better way to solve it?
After putting v1, v2, v3, v4 into a matrix and obtain its row echelon form, obtain the linear independent set from the corresponding pivotal ones. Suppose v1, v2 and v3 are linearly independent. Then just choose u1=v1, u2=v2 and u3=v3 by assigning appropriate values for ai's.
After putting v1, v2, v3, v4 into a matrix and obtain its row echelon form, obtain the linear independent set from the corresponding pivotal ones. Suppose v1, v2 and v3 are linearly independent. Then just choose u1=v1, u2=v2 and u3=v3 by assigning appropriate values for ai's.