2i + 6j - 1k ---- a
 2i - 2j + 4k ---- 2b
-3i - 4j - 1k ---- -c
-------------------------- Add
 1i + 0j + 2k
= i + 2k
=======================

I think you'll find that it's easier to use the < p,q,r > notation in
vector calculus than the pi+qj+rk notation.  Otherwise, you are likely
to get letters used for vectors and letters used for scalars confused.
Ordinary letters are used for scalars and letters in bold-face italics are used
for vectors.  The < p,q,r > avoids having to use so many bold-face italic
letters.

given that a=2l+3j-k, b=l-j+2k, and c=3l+4j+k, find
(a) a+2b-c

a = < 2,3,-1 >, b = < 1,-1,2 >, c = < 3,4,1 >

a+2b-c = < 2,3,-1 > + 2< 1,-1,2 > - < 3,4,1 > =
         < 2,3,-1 > + < 2,-2,4 > + < -3,-4,-1 > =
         < 2+2-3,3-2-4,-1+4-1 > = < 1,-3,2 > = l-3j+2k

(b) a vector d such that a+b+c+d=0,

Let d = pl+qj+rk = < p,q,r >

a+b+c+d = < 2,3,-1 > + < 1,-1,2 > + < 3,4,1 > + < p,q,r > = 0 = < 0,0,0 >

          < 2+1+3+p,3-1+4+q,-1+2+1+r > = < 0,0,0 >

               < 6+p,6+q,2+r > = < 0,0,0 >

                6+p=0;   6+q=0;   2+r=0
                  p=-6;    q=-6;    r=-2

d = pl+qj+rk = < p,q,r > = < -6,-6,-2 >

and (c) a vector d such that a-b+c+3d=0

Let d = pl+qj+rk = < p,q,r >

a-b+c+3d = < 2,3,-1 > - < 1,-1,2 > + < 3,4,1 > + 3< p,q,r > = 0 = < 0,0,0 >
         = < 2,3,-1 > + < -1,1,-2 > + < 3,4,1 > + < 3p,3q,3r > = 0 = < 0,0,0 >
          < 2-1+3+3p,3+1+4+3q,-1-2+1+3r > = < 0,0,0 >
           < 4+3p,8+3q,-3+3r > = < 0,0,0 >

                4+3p=0;   8+3q=0;   -2+3r=0
                  3p=-4;    3q=-8;     3r=2
                   p=-4/3;   q=-8/3;    r=2/3

d = pl+qj+rk = < p,q,r > = < -4/3,-8/3,2/3 >

Edwin