1. **Problem statement:** Find the distance between the two lines $l_1$ and $l_2$ given by: $l_1:$ $x = t, y = t, z = t, t \in \mathbb{R}$ $l_2:$ $x = 1, y = 2, z = t, t \in \mathbb{R}$ 2. **Formula and approach:** The distance $d$ between two skew lines can be found using the formula: $$d = \frac{|(\mathbf{a_2} - \mathbf{a_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2})|}{|\mathbf{b_1} \times \mathbf{b_2}|}$$ where $\mathbf{a_1}$ and $\mathbf{a_2}$ are position vectors of points on lines $l_1$ and $l_2$, and $\mathbf{b_1}$ and $\mathbf{b_2}$ are direction vectors of $l_1$ and $l_2$. 3. **Identify vectors:** - For $l_1$, direction vector $\mathbf{b_1} = (1,1,1)$ (since $x=y=z=t$) - For $l_2$, direction vector $\mathbf{b_2} = (0,0,1)$ (since $x=1, y=2$ fixed, $z=t$) - Choose point on $l_1$: $\mathbf{a_1} = (0,0,0)$ at $t=0$ - Choose point on $l_2$: $\mathbf{a_2} = (1,2,0)$ at $t=0$ 4. **Calculate cross product $\mathbf{b_1} \times \mathbf{b_2}$:** $$\mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 1 \\ 0 & 0 & 1 \end{vmatrix} = (1 \cdot 1 - 1 \cdot 0)\mathbf{i} - (1 \cdot 1 - 1 \cdot 0)\mathbf{j} + (1 \cdot 0 - 1 \cdot 0)\mathbf{k} = (1, -1, 0)$$ 5. **Calculate magnitude of cross product:** $$|\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{2}$$ 6. **Calculate vector between points:** $$\mathbf{a_2} - \mathbf{a_1} = (1, 2, 0) - (0, 0, 0) = (1, 2, 0)$$ 7. **Calculate dot product:** $$|(\mathbf{a_2} - \mathbf{a_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2})| = |(1, 2, 0) \cdot (1, -1, 0)| = |1 \cdot 1 + 2 \cdot (-1) + 0 \cdot 0| = |1 - 2| = 1$$ 8. **Calculate distance:** $$d = \frac{1}{\sqrt{2}}$$ **Final answer:** The distance between the two lines is $\frac{1}{\sqrt{2}}$.