In a triangle $$ABC$$, right angle at vertex $$A$$, if the position vectors of $$A$$, $$B$$ and $$C$$ are respectively $$3\hat{i} + \hat{j} - \hat{k}$$, $$-\hat{i} + 3\hat{j} + p\hat{k}$$ and $$5\hat{i} + q\hat{j} - 4\hat{k}$$, then the point $$(p, q)$$ lies on a line:

In triangle $$ABC$$, the right angle is at vertex $$A$$. The position vectors are given as:

Position vector of $$A$$: $$3\hat{i} + \hat{j} - \hat{k}$$

Position vector of $$B$$: $$-\hat{i} + 3\hat{j} + p\hat{k}$$

Position vector of $$C$$: $$5\hat{i} + q\hat{j} - 4\hat{k}$$

Since the angle at $$A$$ is right-angled, the vectors $$\overrightarrow{AB}$$ and $$\overrightarrow{AC}$$ are perpendicular. Therefore, their dot product is zero.

First, compute $$\overrightarrow{AB}$$:

$$\overrightarrow{AB} = \text{Position vector of } B - \text{Position vector of } A$$

$$\overrightarrow{AB} = (-\hat{i} + 3\hat{j} + p\hat{k}) - (3\hat{i} + \hat{j} - \hat{k})$$

$$\overrightarrow{AB} = -\hat{i} + 3\hat{j} + p\hat{k} - 3\hat{i} - \hat{j} + \hat{k}$$

$$\overrightarrow{AB} = (-\hat{i} - 3\hat{i}) + (3\hat{j} - \hat{j}) + (p\hat{k} + \hat{k})$$

$$\overrightarrow{AB} = -4\hat{i} + 2\hat{j} + (p + 1)\hat{k}$$

Next, compute $$\overrightarrow{AC}$$:

$$\overrightarrow{AC} = \text{Position vector of } C - \text{Position vector of } A$$

$$\overrightarrow{AC} = (5\hat{i} + q\hat{j} - 4\hat{k}) - (3\hat{i} + \hat{j} - \hat{k})$$

$$\overrightarrow{AC} = 5\hat{i} + q\hat{j} - 4\hat{k} - 3\hat{i} - \hat{j} + \hat{k}$$

$$\overrightarrow{AC} = (5\hat{i} - 3\hat{i}) + (q\hat{j} - \hat{j}) + (-4\hat{k} + \hat{k})$$

$$\overrightarrow{AC} = 2\hat{i} + (q - 1)\hat{j} - 3\hat{k}$$

Since $$\overrightarrow{AB}$$ and $$\overrightarrow{AC}$$ are perpendicular, their dot product is zero:

$$\overrightarrow{AB} \cdot \overrightarrow{AC} = 0$$

Substitute the vectors:

$$[-4\hat{i} + 2\hat{j} + (p + 1)\hat{k}] \cdot [2\hat{i} + (q - 1)\hat{j} - 3\hat{k}] = 0$$

Compute the dot product component-wise:

$$(-4) \times 2 + 2 \times (q - 1) + (p + 1) \times (-3) = 0$$

Simplify:

$$-8 + 2(q - 1) - 3(p + 1) = 0$$

Expand the terms:

$$-8 + 2q - 2 - 3p - 3 = 0$$

Combine like terms:

$$-8 - 2 - 3 + 2q - 3p = 0$$

$$-13 + 2q - 3p = 0$$

Rearrange to:

$$2q - 3p = 13$$

This equation represents a straight line in the $$pq$$-plane. Rewrite it in slope-intercept form:

$$2q = 3p + 13$$

$$q = \frac{3}{2}p + \frac{13}{2}$$

The slope of this line is $$\frac{3}{2}$$, which is positive.

A positive slope indicates that the line makes an acute angle with the positive direction of the $$x$$-axis.

Now, evaluate the options:

A. Making an obtuse angle with the positive direction of $$x$$-axis: This requires a negative slope, but the slope is positive, so false.

B. Parallel to $$x$$-axis: This requires a slope of zero, but the slope is $$\frac{3}{2} \neq 0$$, so false.

C. Parallel to $$y$$-axis: This requires an undefined slope (vertical line), but the slope is defined, so false.

D. Making an acute angle with the positive direction of $$x$$-axis: Since the slope is positive, this is true.

Hence, the correct answer is Option D.