1. **Problem statement:** Given vectors $\mathbf{AB} = \mathbf{p}$, $\mathbf{CA} = \mathbf{q}$, and $\mathbf{DC} = 3\mathbf{AB} = 3\mathbf{p}$, we need to: (a) Express $\mathbf{DA}$ in terms of $\mathbf{p}$ and $\mathbf{q}$. (b) Given point $E$ such that $\mathbf{BE} = k\mathbf{q}$: (i) Name the special quadrilateral $ACBE$. (ii) Express $\mathbf{AE}$ in terms of $\mathbf{p}$, $\mathbf{q}$, and $k$. (iii) Given points $D$, $A$, and $E$ are collinear, find $k$. 2. **Part (a): Express $\mathbf{DA}$** We know $\mathbf{DA} = \mathbf{DC} + \mathbf{CA}$ by vector addition. Given $\mathbf{DC} = 3\mathbf{p}$ and $\mathbf{CA} = \mathbf{q}$, $$\mathbf{DA} = 3\mathbf{p} + \mathbf{q}.$$ 3. **Part (b)(i): Name of quadrilateral $ACBE$** Since $E$ lies such that $\mathbf{BE} = k\mathbf{q}$, and $\mathbf{CA} = \mathbf{q}$, points $A$, $C$, $B$, and $E$ form a parallelogram because opposite sides are parallel and equal in vector terms. **Answer:** $ACBE$ is a parallelogram. 4. **Part (b)(ii): Express $\mathbf{AE}$** Vector $\mathbf{AE} = \mathbf{AB} + \mathbf{BE}$. Given $\mathbf{AB} = \mathbf{p}$ and $\mathbf{BE} = k\mathbf{q}$, $$\mathbf{AE} = \mathbf{p} + k\mathbf{q}.$$ 5. **Part (b)(iii): Find $k$ given $D$, $A$, and $E$ are collinear** Collinearity means vectors $\mathbf{DA}$ and $\mathbf{AE}$ are scalar multiples: $$\mathbf{DA} = \mathbf{AE}.$$ From above, $$3\mathbf{p} + \mathbf{q} = \mathbf{p} + k\mathbf{q}.$$ Rearranging, $$3\mathbf{p} - \mathbf{p} = k\mathbf{q} - \mathbf{q}$$ $$2\mathbf{p} = (k - 1)\mathbf{q}.$$ Since $\mathbf{p}$ and $\mathbf{q}$ are independent vectors, the only way this holds is if both sides are zero or proportional. For this to be true, $\mathbf{p}$ must be parallel to $\mathbf{q}$, or the coefficients must be zero. Assuming $\mathbf{p}$ and $\mathbf{q}$ are independent, the only solution is when both sides equal zero, which is impossible here. So $\mathbf{p}$ and $\mathbf{q}$ must be parallel. If $\mathbf{p} = m\mathbf{q}$ for some scalar $m$, then: $$2m\mathbf{q} = (k - 1)\mathbf{q} \implies 2m = k - 1 \implies k = 2m + 1.$$ If $\mathbf{p}$ and $\mathbf{q}$ are not parallel, no solution exists. **Final answers:** (a) $\boxed{\mathbf{DA} = 3\mathbf{p} + \mathbf{q}}$ (b)(i) $ACBE$ is a parallelogram. (b)(ii) $\boxed{\mathbf{AE} = \mathbf{p} + k\mathbf{q}}$ (b)(iii) If $\mathbf{p} = m\mathbf{q}$, then $\boxed{k = 2m + 1}$; otherwise, no solution for $k$ exists.