Let $$S$$ denote the locus of the point of intersection of the pair of lines

$$4x - 3y = 12\alpha$$,

$$4\alpha x + 3\alpha y = 12$$,

where $$\alpha$$ varies over the set of non-zero real numbers. Let $$T$$ be the tangent to $$S$$ passing through the points $$(p, 0)$$ and $$(0, q)$$, $$q > 0$$, and parallel to the line $$4x - \frac{3}{\sqrt{2}} y = 0$$.

Then the value of $$pq$$ is

Let the two concurrent lines be$$4x-3y=12\alpha \qquad -(1)$$and$$4\alpha x+3\alpha y=12 \qquad -(2)$$with $$\alpha\neq 0$$.

Coordinates of the intersection
From $$(1)$$ we get $$4x-3y=12\alpha$$.
Divide $$(2)$$ by $$\alpha$$ to obtain $$4x+3y=\dfrac{12}{\alpha}\qquad -(3)$$.

Add $$(1)$$ and $$(3)$$:
$$8x=12\alpha+\dfrac{12}{\alpha}\;\Longrightarrow\;x=\dfrac{3}{2}\left(\alpha+\dfrac{1}{\alpha}\right).$$

Subtract $$(1)$$ from $$(3)$$:
$$6y=\dfrac{12}{\alpha}-12\alpha\;\Longrightarrow\;y=2\left(\dfrac{1}{\alpha}-\alpha\right).$$

Eliminating $$\alpha$$ to get the locus $$S$$
Let $$t=\alpha+\dfrac{1}{\alpha}=\dfrac{2}{3}x$$ and$$s=\alpha-\dfrac{1}{\alpha}=-\dfrac{y}{2}.$$ For any real $$\alpha\neq 0$$ we have $$(\alpha+\tfrac{1}{\alpha})^{2}-(\alpha-\tfrac{1}{\alpha})^{2}=4.$$ Substituting $$t$$ and $$s$$, $$\left(\dfrac{2}{3}x\right)^{2}-\left(-\dfrac{y}{2}\right)^{2}=4.$$ This simplifies to $$\dfrac{4}{9}x^{2}-\dfrac{y^{2}}{4}=4 \;\Longrightarrow\; \dfrac{x^{2}}{9}-\dfrac{y^{2}}{16}=1.$$ Hence $$S$$ is the hyperbola $$\boxed{\dfrac{x^{2}}{9}-\dfrac{y^{2}}{16}=1}.$

Equation of the required tangent $$T$$
The tangent is parallel to the line $$4x-\dfrac{3}{$$\sqrt$$2}y=0$$ whose slope is $$m=\dfrac{4$$\sqrt$$2}{3}.$$
Let the tangent cut the axes at $$(p,0)$$ and $$(0,q)\;(q\gt 0).$$ In intercept form it is $$\dfrac{x}{p}+\dfrac{y}{q}=1,$$ whose slope is $$-\dfrac{q}{p}.$$ Parallelism gives $$-\dfrac{q}{p}=m=\dfrac{4$$\sqrt$$2}{3}\;\Longrightarrow\;q=-mp.$$

Tangency condition
For the hyperbola $$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$$ with $$a^{2}=9,\;b^{2}=16,$$ the line $$y=mx+c$$ is a tangent iff $$c^{2}=a^{2}m^{2}-b^{2}.$$ Here $$y=mx+c$$ with $$c=-mp,$$ so $$c^{2}=(-mp)^{2}=m^{2}p^{2}.$$ Using the tangency criterion, $$m^{2}p^{2}=a^{2}m^{2}-b^{2} \;\Longrightarrow\; p^{2}=a^{2}-\dfrac{b^{2}}{m^{2}}.$$

Compute each term:
$$a^{2}m^{2}=9$$\left$$(\dfrac{4$$\sqrt$$2}{3}$$\right$$)^{2}=9$$\cdot$$\dfrac{32}{9}=32,$$ and therefore $$c^{2}=a^{2}m^{2}-b^{2}=32-16=16\quad$$\Rightarrow$$\quad c=4 \;(c\gt 0).$$

Since $$c=-mp,$$ we have $$4=-mp\;\Longrightarrow\;p=-\dfrac{4}{m} =-\dfrac{4}{\dfrac{4$$\sqrt$$2}{3}}=-\dfrac{3}{$$\sqrt$$2}.$$ Also $$q=-mp=-$$\left$$(\dfrac{4$$\sqrt$$2}{3}$$\right$$)\!$$\left$$(-\dfrac{3}{$$\sqrt$$2}$$\right$$)=4.$$

Required product
$$pq=$$\left$$(-\dfrac{3}{$$\sqrt$$2}$$\right$$)(4)=-\dfrac{12}{$$\sqrt$$2}=-6$$\sqrt$$2.$$

Thus the correct value is $$\boxed{-6$$\sqrt$$2}$$, which corresponds to
Option A.