Vectors

$\sqrt{4^2 + (-3)^2} = \sqrt{16+9} = \sqrt{25} = 5$. The negative sign disappears when squared.

Add component by component: $2+3 = 5$ and $1 + (-4) = -3$.

$\overrightarrow{AB} = (7-2)\mathbf{i} + (11-(-1))\mathbf{j} = 5\mathbf{i} + 12\mathbf{j}$. Then $\sqrt{25 + 144} = 13$.

The magnitude is $\sqrt{36+64} = 10$, so divide each component by 10.

$\mathbf{a} + \lambda\mathbf{b} = (3+\lambda)\mathbf{i} + (2+4\lambda)\mathbf{j}$. Parallel to $\mathbf{i}$ means there is no $\mathbf{j}$ component, so $2 + 4\lambda = 0$ and $\lambda = -\tfrac12$.

$\overrightarrow{AB} = 3\mathbf{i}+6\mathbf{j}$ and $\overrightarrow{BC} = 2\mathbf{i}+4\mathbf{j}$. Since $\overrightarrow{BC} = \tfrac23\overrightarrow{AB}$, the two vectors are parallel; and they both pass through $B$, so the three points lie on one straight line.