Three Dimensional Geometry

The line passing through \((1,1,-1)\) and parallel to the vector \(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}\) meets the line \(\frac{x-3}{-1}=\frac{y+2}{5}=\frac{z-2}{-4}\) at \(A\) and the plane \(2 x-y+2 z+7=0\) at \(B\). Then \(A B\) is equal to