The direction ratios of the two lines are \((1,-2,3)\) and \((2,-1,-1)\). The normal to the plane is perpendicular to both, so it's their cross product: \(\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & -1 & -1 \end{vmatrix} = \hat{i}(2-(-3)) - \hat{j}(-1-6) + \hat{k}(-1-(-4)) = 5\hat{i} + 7\hat{j} + 3\hat{k}\). The plane equation through \((1,-1,-1)\) with normal \((5,7,3)\) is \(5(x-1) + 7(y+1) + 3(z+1) = 0\), which gives \(5x+7y+3z+5 = 0\). Distance from \((1,3,-7)\) is \(\frac{|5(1)+7(3)+3(-7)+5|}{\sqrt{25+49+9}} = \frac{|5+21-21+5|}{\sqrt{83}} = \frac{10}{\sqrt{83}}\).