Let the position vectors of two points $$P$$ and $$Q$$ be $$3\hat{i} - \hat{j} + 2\hat{k}$$ and $$\hat{i} + 2\hat{j} - 4\hat{k}$$, respectively. Let $$R$$ and $$S$$ be two points such that the direction ratios of lines $$PR$$ and $$QS$$ are $$(4, -1, 2)$$ and $$(-2, 1, -2)$$, respectively. Let lines $$PR$$ and $$QS$$ intersect at $$T$$. If the vector $$\vec{TA}$$ is perpendicular to both $$\vec{PR}$$ and $$\vec{QS}$$ and the length of vector $$\vec{TA}$$ is $$\sqrt{5}$$ units, then the modulus of a position vector of $$A$$ is:

We are given $$P = (3, -1, 2)$$ and $$Q = (1, 2, -4)$$. Line $$PR$$ has direction ratios $$(4, -1, 2)$$ and line $$QS$$ has direction ratios $$(-2, 1, -2)$$.

The parametric equations of line $$PR$$ are: $$x = 3 + 4t,\; y = -1 - t,\; z = 2 + 2t$$.

The parametric equations of line $$QS$$ are: $$x = 1 - 2s,\; y = 2 + s,\; z = -4 - 2s$$.

At the intersection point $$T$$, we equate coordinates: $$3 + 4t = 1 - 2s$$, $$-1 - t = 2 + s$$, and $$2 + 2t = -4 - 2s$$.

From the second equation: $$t + s = -3$$, so $$s = -3 - t$$. Substituting into the first equation: $$3 + 4t = 1 - 2(-3 - t) = 7 + 2t$$, giving $$2t = 4$$, so $$t = 2$$ and $$s = -5$$.

Verification with the third equation: $$2 + 4 = 6$$ and $$-4 - 2(-5) = 6$$. Confirmed.

So $$T = (3 + 8, -1 - 2, 2 + 4) = (11, -3, 6)$$.

The vector $$\vec{TA}$$ is perpendicular to both $$\vec{PR} = (4, -1, 2)$$ and $$\vec{QS} = (-2, 1, -2)$$. We compute the cross product: $$\vec{PR} \times \vec{QS} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 2 \\ -2 & 1 & -2 \end{vmatrix} = \hat{i}(2 - 2) - \hat{j}(-8 + 4) + \hat{k}(4 - 2) = (0, 4, 2)$$.

The unit vector along this direction is $$\frac{(0, 4, 2)}{\sqrt{0 + 16 + 4}} = \frac{(0, 4, 2)}{\sqrt{20}} = \frac{(0, 4, 2)}{2\sqrt{5}}$$.

Since $$|\vec{TA}| = \sqrt{5}$$, we have $$\vec{TA} = \pm \sqrt{5} \cdot \frac{(0, 4, 2)}{2\sqrt{5}} = \pm (0, 2, 1)$$.

The position vector of $$A$$ is $$T + \vec{TA} = (11, -3 \pm 2, 6 \pm 1)$$, giving $$A = (11, -1, 7)$$ or $$A = (11, -5, 5)$$.

For $$A = (11, -1, 7)$$: $$|A| = \sqrt{121 + 1 + 49} = \sqrt{171}$$.

For $$A = (11, -5, 5)$$: $$|A| = \sqrt{121 + 25 + 25} = \sqrt{171}$$.

In both cases, $$|A| = \sqrt{171}$$, which corresponds to option (2).