If
$A=3i+2j-3k$
&B= 2i-2j-k$
$C=2i-j-3k$
What is $2C-[2A.(2B+4C)]B?$

We will be breaking down the solution one step at a time. First we begin with
$2A=2(3i+2j-3k)=6i+4j-6k$
$2B=2(2i-2j-k)=4i-4j-2k$
$2C=2(2i-j-3k)=8i-4j-12k$
$4C=4(2i-j-3k)=8i-4j-12k$
Now we solve for
$2B+4C$
$2B+4C=(4i-4j-2k)+(8i-4j-12k)=12i-8j-14k$

Here, i am using the rule for scaler multiplication
$2A.(2B+4C)=(6i+4j-6k)(12i-8j-14k)=72-32+84=124$
Solving For 
$[2A.(2B+4C)]B$
 $[124](2i-2j-k)=248i-248j-124k$

I will equally use scaler multiplication for
$2C-[2A.(2B+4C)]B$
$(4i-2j-6k)-(248i-248j-124k)$
$=- 244i+246j+118k$
$=2[126i-123j+59k]$