Let P be a point in the plane of the vectors $$ \overrightarrow{AB}=3\widehat{i} + \widehat{j}-\widehat{k} \text{ and }\overrightarrow{AC}=\widehat{i}-\widehat{j}+3\widehat{k}$$ such that P is equidistant from the Lines AB and AC. If $$ \mid \overrightarrow{AP} \mid=\frac{\sqrt{5}}{2} $$ then the area of the triangle ABP is:

The vectors are given as $$\overrightarrow{AB} = 3\hat{i} + \hat{j} - \hat{k}$$ and $$\overrightarrow{AC} = \hat{i} - \hat{j} + 3\hat{k}$$. Point A is taken as the origin, so $$\vec{A} = \vec{0}$$, $$\vec{B} = 3\hat{i} + \hat{j} - \hat{k}$$, and $$\vec{C} = \hat{i} - \hat{j} + 3\hat{k}$$. Point P lies in the plane of $$\vec{B}$$ and $$\vec{C}$$, so $$\overrightarrow{AP} = \vec{P} = x \vec{B} + y \vec{C}$$ for scalars $$x$$ and $$y$$. Thus,

$$\vec{P} = x(3\hat{i} + \hat{j} - \hat{k}) + y(\hat{i} - \hat{j} + 3\hat{k}) = (3x + y)\hat{i} + (x - y)\hat{j} + (-x + 3y)\hat{k}$$

Given that P is equidistant from lines AB and AC, and lines AB and AC pass through A with direction vectors $$\vec{B}$$ and $$\vec{C}$$, respectively, the distance from P to line AB is $$\frac{|\vec{P} \times \vec{B}|}{|\vec{B}|}$$ and to line AC is $$\frac{|\vec{P} \times \vec{C}|}{|\vec{C}|}$$. Equating the distances:

$$\frac{|\vec{P} \times \vec{B}|}{|\vec{B}|} = \frac{|\vec{P} \times \vec{C}|}{|\vec{C}|}$$

First, compute $$|\vec{B}| = \sqrt{3^2 + 1^2 + (-1)^2} = \sqrt{11}$$ and $$|\vec{C}| = \sqrt{1^2 + (-1)^2 + 3^2} = \sqrt{11}$$. Since $$|\vec{B}| = |\vec{C}|$$, the equation simplifies to:

$$|\vec{P} \times \vec{B}| = |\vec{P} \times \vec{C}|$$

Squaring both sides:

$$|\vec{P} \times \vec{B}|^2 = |\vec{P} \times \vec{C}|^2$$

Using the identity $$|\vec{U} \times \vec{V}|^2 = |\vec{U}|^2 |\vec{V}|^2 - (\vec{U} \cdot \vec{V})^2$$,

$$|\vec{P}|^2 |\vec{B}|^2 - (\vec{P} \cdot \vec{B})^2 = |\vec{P}|^2 |\vec{C}|^2 - (\vec{P} \cdot \vec{C})^2$$

Substituting $$|\vec{B}| = |\vec{C}| = \sqrt{11}$$:

$$11|\vec{P}|^2 - (\vec{P} \cdot \vec{B})^2 = 11|\vec{P}|^2 - (\vec{P} \cdot \vec{C})^2$$

Thus,

$$(\vec{P} \cdot \vec{B})^2 = (\vec{P} \cdot \vec{C})^2$$

So,

$$\vec{P} \cdot \vec{B} = \vec{P} \cdot \vec{C} \quad \text{or} \quad \vec{P} \cdot \vec{B} = -\vec{P} \cdot \vec{C}$$

Also, $$|\overrightarrow{AP}| = \frac{\sqrt{5}}{2}$$, so $$|\vec{P}|^2 = \left(\frac{\sqrt{5}}{2}\right)^2 = \frac{5}{4}$$.

Compute $$\vec{P} \cdot \vec{B}$$ and $$\vec{P} \cdot \vec{C}$$:

$$\vec{P} \cdot \vec{B} = (3x + y)(3) + (x - y)(1) + (-x + 3y)(-1) = 9x + 3y + x - y + x - 3y = 11x - y$$

$$\vec{P} \cdot \vec{C} = (3x + y)(1) + (x - y)(-1) + (-x + 3y)(3) = 3x + y - x + y - 3x + 9y = -x + 11y$$

Set the conditions:

Case 1: $$11x - y = -x + 11y$$
Solving: $$11x - y = -x + 11y \implies 12x = 12y \implies x = y$$

Case 2: $$11x - y = -(-x + 11y) = x - 11y$$
Solving: $$11x - y = x - 11y \implies 10x = -10y \implies y = -x$$

Now, $$|\vec{P}|^2 = (3x + y)^2 + (x - y)^2 + (-x + 3y)^2 = 11x^2 + 11y^2 - 2xy = \frac{5}{4}$$.

For Case 1 ($$x = y$$):
Substitute $$y = x$$:
$$11x^2 + 11x^2 - 2x(x) = 22x^2 - 2x^2 = 20x^2 = \frac{5}{4} \implies x^2 = \frac{5}{80} = \frac{1}{16} \implies x = \pm \frac{1}{4}, y = \pm \frac{1}{4}$$

For $$x = y = \frac{1}{4}$$:
$$\vec{P} = \left(3 \cdot \frac{1}{4} + \frac{1}{4}\right)\hat{i} + \left(\frac{1}{4} - \frac{1}{4}\right)\hat{j} + \left(-\frac{1}{4} + 3 \cdot \frac{1}{4}\right)\hat{k} = \hat{i} + 0\hat{j} + \frac{1}{2}\hat{k}$$

For $$x = y = -\frac{1}{4}$$:
$$\vec{P} = \left(3 \cdot -\frac{1}{4} + -\frac{1}{4}\right)\hat{i} + \left(-\frac{1}{4} - -\frac{1}{4}\right)\hat{j} + \left(-\left(-\frac{1}{4}\right) + 3 \cdot -\frac{1}{4}\right)\hat{k} = -\hat{i} + 0\hat{j} - \frac{1}{2}\hat{k}$$

For Case 2 ($$y = -x$$):
Substitute $$y = -x$$:
$$11x^2 + 11(-x)^2 - 2x(-x) = 11x^2 + 11x^2 + 2x^2 = 24x^2 = \frac{5}{4} \implies x^2 = \frac{5}{96} \implies x = \pm \frac{\sqrt{30}}{24}, y = \mp \frac{\sqrt{30}}{24}$$

The area of triangle ABP is given by $$\frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AP}| = \frac{1}{2} |\vec{B} \times \vec{P}|$$.

For Case 1 with $$\vec{P} = \hat{i} + \frac{1}{2}\hat{k}$$:
$$\vec{B} \times \vec{P} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -1 \\ 1 & 0 & \frac{1}{2} \end{vmatrix} = \hat{i}\left(1 \cdot \frac{1}{2} - (-1) \cdot 0\right) - \hat{j}\left(3 \cdot \frac{1}{2} - (-1) \cdot 1\right) + \hat{k}\left(3 \cdot 0 - 1 \cdot 1\right) = \frac{1}{2}\hat{i} - \frac{5}{2}\hat{j} - \hat{k}$$
Magnitude: $$\left|\vec{B} \times \vec{P}\right| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{5}{2}\right)^2 + (-1)^2} = \sqrt{\frac{1}{4} + \frac{25}{4} + 1} = \sqrt{\frac{30}{4}} = \frac{\sqrt{30}}{2}$$
Area: $$\frac{1}{2} \times \frac{\sqrt{30}}{2} = \frac{\sqrt{30}}{4}$$

Similarly, for $$\vec{P} = -\hat{i} - \frac{1}{2}\hat{k}$$, the magnitude is the same, so the area is also $$\frac{\sqrt{30}}{4}$$.

For Case 2, the area calculation yields $$\frac{5}{4}$$, which is not among the options. Only Case 1 gives an area matching option A.

Thus, the area of triangle ABP is $$\frac{\sqrt{30}}{4}$$.