$\text{Let}\mathrm{ S }\text{denote the locus of the point of intersection of the pair of lines}$
$4x-3y=12\alpha ,$
$4\alpha x+3\alpha y=12,$
$\text{where }\alpha \text{ varies over the set of non-zero real numbers. Let}\mathrm{ T}\text{ be the tangent to}$

$S\text{ passing through the points}(p,0)\text{and}(0,q),p>0,\text{and parallel to the line}\mathrm{ 4}x-\frac{3}{\sqrt{2}}y=0.$

Then the value of pq is

The correct answer is $-6\sqrt{2}$.

Step 1: Find the locus $S$ by eliminating $\alpha$
We are given the system of equations of the intersecting lines:
(1) $4x-3y=12\alpha$
(2) $4\alpha x+3\alpha y=12$

From equation (2), since $\alpha$ is a non-zero real number:
$\alpha (4x+3y)=12\Rightarrow \alpha =\frac{12}{4x+3y}$

Substituting this expression for $\alpha$ into equation (1) gives:
$4x-3y=12\left(\frac{12}{4x+3y}\right)$

Multiplying both sides by $4x+3y$:
$(4x-3y)(4x+3y)=144$
$16x^2-9y^2=144$

Dividing both sides by 144 yields the standard equation of a hyperbola:
$\frac{x^2}{9}-\frac{y^2}{16}=1$
Thus, the locus $S$ is a hyperbola with parameters:
$a^2=9$ and $b^2=16$

Step 2: Find the equation of the tangent line $T$
The tangent $T$ is parallel to the line:
$4x-\frac{3}{\sqrt{2}}y=0$
The slope $m$ of this line is:
$m=\frac{4}{3/\sqrt{2}}=\frac{4\sqrt{2}}{3}$

The equation of a tangent to a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ with slope $m$ is:
$y=mx\pm \sqrt{a^2{m}^2-b^2}$

First, compute the term inside the square root:
$a^2{m}^2-b^2=9·{\left(\frac{4\sqrt{2}}{3}\right)}^2-16=9·\frac{32}{9}-16=32-16=16$

Thus, the equation of the tangent is:
$y=\frac{4\sqrt{2}}{3}x\pm 4$

Step 3: Find the values of $p$ and $q$
The tangent passes through the x-intercept $(p,0)$ and the y-intercept .
For the x-intercept:
$0=\frac{4\sqrt{2}}{3}p\pm 4\Rightarrow p=\mp \frac{3}{\sqrt{2}}$
Since we are given that $p>0$, we must choose the positive sign:
$p=\frac{3}{\sqrt{2}}$

This corresponds to the tangent line:
$y=\frac{4\sqrt{2}}{3}x-4$
The y-intercept of this line is:
$q=-4$

Step 4: Calculate the product $pq$
$pq=\left(\frac{3}{\sqrt{2}}\right)(-4)=-\frac{12}{\sqrt{2}}=-6\sqrt{2}$