Answer
(a) A basis for the column is
$$S=\left\{\left[\begin{array}{rrrr}{2} \\ {7} \\ {-2} \\ {2} \end{array}\right],\left[\begin{array}{rrrr} {-3} \\ {-6} \\ {1} \\ {-2} \end{array}\right]\right\}.$$
(b) The rank of the matrix is $2$.
Work Step by Step
Given the matrix
$$
\left[\begin{array}{rrrr}{2} & {4} & {-3} & {-6} \\ {7} & {14} & {-6} & {-3} \\ {-2} & {-4} & {1} & {-2} \\ {2} & {4} & {-2} & {-2}\end{array}\right].
$$
The reduced row echelon form is given by
$$
\left[ \begin {array}{cccc} 1&2&0&3\\ 0&0&1&4
\\ 0&0&0&0\\ 0&0&0&0\end {array}
\right]
.
$$
(a) A basis for the column space are the columns corresponding to the columns that have the leading 1's. That is,
$$S=\left\{\left[\begin{array}{rrrr}{2} \\ {7} \\ {-2} \\ {2} \end{array}\right],\left[\begin{array}{rrrr} {-3} \\ {-6} \\ {1} \\ {-2} \end{array}\right]\right\}.$$
(b) The rank of the matrix is $2$.