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]