Answer
$$-a+b+6c$$
Work Step by Step
We're given:
$$-3(a-c)+2(a+2b)+3(c-b)$$
Using the distributivity of scalar-vector multiplication over vector addition (Theorem 1.1.e), we have:
$$(-3)a+(-3)(-1)c+(2)a+(2)(2)b+(3)c+(3)(-1)b$$
Through scalar multiplication (Theorem 1.1.g), we get:
$$(-3)a+(3)c+(2)a+(4)b+(3)c+(-3)b$$
Then, rearrange using the commutativity of vector addition (Theorem 1.1.a)
$$(-3)a+(2)a+(4)b+(-3)b+(3)c+(3)c$$
Using the distributivity of scalar-vector multiplication over scalar addition, we get:
$$(-3+2)a+(4-3)b+(3+3)c$$
Lastly, we simplify by scalar addition to get:
$$-a+b+6c$$
