1. **Problem statement:** (i) Find the equation of the line through points $[1,2,3]$ and $[1,1,1]$ in $\mathbb{RP}^2$. (ii) Find the intersection point of lines $x + 2y + 3z = 0$ and $x + y + z = 0$ in $\mathbb{RP}^2$. 2. **Recall:** In projective plane $\mathbb{RP}^2$, points are given by homogeneous coordinates $[a,b,c]$. Two points $[a,b,c]$ and $[a',b',c']$ represent the same point if there exists $t \neq 0$ such that $a = ta', b = tb', c = tc'$. 3. **Line through two points:** The line through points $P = [a,b,c]$ and $Q = [a',b',c']$ is given by the cross product: $$L = P \times Q = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a & b & c \\ a' & b' & c' \end{vmatrix}$$ This yields the line coefficients $[l,m,n]$ such that $lx + my + nz = 0$. 4. **Calculate line through $[1,2,3]$ and $[1,1,1]$:** $$L = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 1 & 1 & 1 \end{vmatrix} = \mathbf{i}(2 \cdot 1 - 3 \cdot 1) - \mathbf{j}(1 \cdot 1 - 3 \cdot 1) + \mathbf{k}(1 \cdot 1 - 2 \cdot 1)$$ $$= \mathbf{i}(2 - 3) - \mathbf{j}(1 - 3) + \mathbf{k}(1 - 2) = [-1, 2, -1]$$ 5. **Equation of the line:** $$-1 \cdot x + 2 \cdot y - 1 \cdot z = 0$$ or equivalently $$-x + 2y - z = 0$$ 6. **Intersection of two lines:** Given lines $L_1: x + 2y + 3z = 0$ and $L_2: x + y + z = 0$, their intersection point $P$ is given by the cross product of their coefficient vectors: $$P = L_1 \times L_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 1 & 1 & 1 \end{vmatrix}$$ 7. **Calculate intersection point:** $$P = \mathbf{i}(2 \cdot 1 - 3 \cdot 1) - \mathbf{j}(1 \cdot 1 - 3 \cdot 1) + \mathbf{k}(1 \cdot 1 - 2 \cdot 1) = [-1, 2, -1]$$ 8. **Intersection point coordinates:** $$P = [-1, 2, -1]$$ **Final answers:** (i) The line through $[1,2,3]$ and $[1,1,1]$ is $$-x + 2y - z = 0$$ (ii) The intersection point of the lines $x + 2y + 3z = 0$ and $x + y + z = 0$ is $$[-1, 2, -1]$$