Let \(\lambda_1, \lambda_2\) be the values of \(\lambda\) for which the points \((1, \lambda, \frac{1}{2})\) and \((-2, 0, 1)\) are at equal distance from the plane \(2x + 3y - 6z + 7 = 0\). If \(\lambda_1 > \lambda_2\), then the distance of the point \((1 - \lambda_2, \lambda_2, \lambda_1)\) from the line \(\frac{x-5}{1} = \frac{y-1}{2} = \frac{z+7}{2}\) is: