Let the required line intersect the given line at parameter \(t\), giving point \((-1-3t, 3+2t, 2-t)\). The direction vector of required line from \((-4,3,1)\) to this point is \((3-3t, 2t, 1-t)\). This must be perpendicular to plane normal \((1,2,-1)\): \((3-3t)(1) + (2t)(2) + (1-t)(-1) = 0\), giving \(3-3t+4t-1+t = 0\), so \(2+2t = 0\) and \(t = -1\). The intersection point is \((2, 1, 3)\). Direction ratios: \((2-(-4), 1-3, 3-1) = (6, -2, 2)\), which simplifies to \((3, -1, 1)\). The line equation is \(\frac{x+4}{3} = \frac{y-3}{-1} = \frac{z-1}{1}\).