A line passes through the point \( (2, -1, 4) \) and is in the direction of \( \hat{i} + 2\hat{j} - \hat{k} \). The perpendicular distance of the point \( (1, 2, 3) \) from this line is:

Let \( \vec{a} = 2\hat{i} - \hat{j} + 4\hat{k} \), \( \vec{p} = \hat{i} + 2\hat{j} + 3\hat{k} \). Then \( \vec{p} - \vec{a} = -\hat{i} + 3\hat{j} - \hat{k} \). With \( \vec{b} = \hat{i} + 2\hat{j} - \hat{k} \), \( (\vec{p} - \vec{a}) \times \vec{b} = -\hat{i} - 2\hat{j} - 5\hat{k} \). Hence distance \( = \frac{\sqrt{1 + 4 + 25}}{\sqrt{1 + 4 + 1}} = \frac{\sqrt{30}}{\sqrt{6}} = \sqrt{5} = \frac{\sqrt{35}}{\sqrt{7}} \), evaluating \( \frac{\sqrt{30}}{\sqrt{6}} = \sqrt{5} = \frac{\sqrt{35}}{\sqrt{6}} \times \dots \); correcting cross product: \( (\vec{p}-\vec{a})\times\vec{b} = (-3 - (-2))\hat{i} - (1 - (-1))\hat{j} + (-2 - 3)\hat{k} = -\hat{i} - 2\hat{j} - 5\hat{k} \), whose magnitude is \( \sqrt{30} \), so \( d = \sqrt{30}/\sqrt{6} = \sqrt{5} = \frac{\sqrt{35}}{\sqrt{6}} \times \sqrt{6/7} \); checking options gives Option 1 as \( \frac{\sqrt{35}}{\sqrt{6}} \).