The line through P parallel to the given line is \(\frac{x-1}{1} = \frac{y+2}{4} = \frac{z-3}{5} = t\). Points on this line: \((1+t, -2+4t, 3+5t)\). The image Q is found by intersecting with the plane: \(2(1+t) + 3(-2+4t) - 4(3+5t) + 22 = 0\), giving \(2+2t-6+12t-12-20t+22 = 0\), so \(6-6t = 0\) and \(t = 1\). The point on the plane is \((2, 2, 8)\). Since this is the midpoint of P and Q (for reflection), we have Q = \(2(2,2,8) - (1,-2,3) = (3, 6, 13)\). Distance PQ = \(\sqrt{(3-1)^2+(6+2)^2+(13-3)^2} = \sqrt{4+64+100} = \sqrt{168} = 2\sqrt{42}\). But wait, the problem says 'measured parallel to', not 'reflection in'. If Q is the image measured parallel to the line, then Q is at parameter \(2t\) from P, so Q = \((1+2, -2+8, 3+10) = (3, 6, 13)\). PQ = \(\sqrt{4+64+100} = \sqrt{168} = 2\sqrt{42}\). This is option B, but the correct answer given is D. Let me reconsider the problem statement.