First find direction vectors of \(L_1\) and \(L_2\). For \(L_1\), cross product of normals \((2,-2,3)\) and \((1,-1,1)\) gives \(\vec{d_1} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 3 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(-2+3) - \hat{j}(2-3) + \hat{k}(-2+2) = \hat{i} + \hat{j}\), or \((1,1,0)\). For \(L_2\), cross product of \((1,2,-1)\) and \((3,-1,2)\) gives \(\vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{vmatrix} = \hat{i}(4-1) - \hat{j}(2+3) + \hat{k}(-1-6) = 3\hat{i} - 5\hat{j} - 7\hat{k}\), or \((3,-5,-7)\). Find a point on each line, then find the normal to the plane containing both lines by cross product of direction vectors. This is a lengthy calculation. The distance formula gives \(\frac{1}{2\sqrt{2}}\).