= Solution
Let $c_1=(1,1,1)^T$ and $c_2=(2,-2,6)^T$. The remaining columns are $3c_1$ and $2c_2$, so these two <vectors> span the <column space>. They are <linearly independent>: the <determinant> of their first two coordinates is $-4\ne0$. Therefore \b[$\{(1,1,1)^T,(2,-2,6)^T\}$ is a <basis> of the range], and the <matrix rank> is two.
Equivalently the range is the plane $y_3=2y_1-y_2$: both displayed basis <vectors> satisfy this equation, and their two-dimensional span fills that plane.
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