Answer
$$AB
=\left[\begin{array}{ccc}{2a_1+a_2-a_3} & {3a_1-a_2+6a_3} & {a_2+4a_3} \end{array}\right] .$$
Work Step by Step
Let $$
A=\left[\begin{array}{ccc}{1} & {0} & {-2} \\ {-3} & {1} & {1} \\ {2} & {0} & {-1}\end{array}\right], B=\left[\begin{array}{ccc}{2} & {3} & {0} \\ {1} & {-1} & {1} \\ {-1} & {6} & {4}\end{array}\right].
$$
Suppose that $A=\left[\begin{array}{ccc}{a_1} & {a_2} & {a_3} \end{array}\right]$, where $a_1,a_2,a_3$ are the columns of $A$.
Now, we have
$$AB=\left[\begin{array}{ccc}{a_1} & {a_2} & {a_3} \end{array}\right]\left[\begin{array}{ccc}{2} & {3} & {0} \\ {1} & {-1} & {1} \\ {-1} & {6} & {4}\end{array}\right]\\
=\left[\begin{array}{ccc}{2a_1+a_2-a_3} & {3a_1-a_2+6a_3} & {a_2+4a_3} \end{array}\right] .$$
