Step 1: Find the equation of the locus S

The point of intersection $(x, y)$ is given by the solution of the two linear equations:

1. $4x - 3y = 12\alpha$
2. $4\alpha x + 3\alpha y = 12$

Our goal is to eliminate the parameter $\alpha$. From equation (1), we can express $\alpha$ in terms of $x$ and $y$: $\alpha = \frac{4x - 3y}{12}$ Since $\alpha$ is a non-zero real number, $4x - 3y \neq 0$.

From equation (2), we can also express $\alpha$: $\alpha (4x + 3y) = 12 \implies \alpha = \frac{12}{4x + 3y}$ Since $\alpha$ is non-zero, $4x + 3y \neq 0$.

Now, we equate the two expressions for $\alpha$: $\frac{4x - 3y}{12} = \frac{12}{4x + 3y}$ Cross-multiplying gives: $(4x - 3y)(4x + 3y) = 12 \times 12$ Using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$: $(4x)^2 - (3y)^2 = 144$ $16x^2 - 9y^2 = 144$ To write this in the standard form of a conic section, we divide the entire equation by 144: $\frac{16x^2}{144} - \frac{9y^2}{144} = 1$ $\frac{x^2}{9} - \frac{y^2}{16} = 1$ This is the equation of a hyperbola. So, the locus S is a hyperbola with $a^2 = 9$ and $b^2 = 16$.

Step 2: Find the equation of the tangent T

The tangent T is parallel to the line $4x - \frac{3}{\sqrt{2}} y = 0$. The slope of this line is given by $m = -\frac{\text{coefficient of } x}{\text{coefficient of } y}$. $m = -\frac{4}{-3/\sqrt{2}} = \frac{4\sqrt{2}}{3}$ Since T is parallel to this line, the slope of tangent T is $m_T = \frac{4\sqrt{2}}{3}$.

The equation of a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with slope $m$ is given by the formula: $y = mx \pm \sqrt{a^2m^2 - b^2}$ Substituting the values $a^2 = 9$, $b^2 = 16$, and $m = \frac{4\sqrt{2}}{3}$: $a^2m^2 - b^2 = 9 \left( \frac{4\sqrt{2}}{3} \right)^2 - 16$ $= 9 \left( \frac{16 \times 2}{9} \right) - 16$ $= 32 - 16 = 16$ So, the condition for tangency is satisfied. The equations of the two possible tangents are: $y = \frac{4\sqrt{2}}{3}x \pm \sqrt{16}$ $y = \frac{4\sqrt{2}}{3}x \pm 4$

Step 3: Identify the correct tangent and find p and q

The tangent T passes through the points $(p, 0)$ and $(0, q)$. This means $p$ is the x-intercept and $q$ is the y-intercept of the tangent line.

From the two possible tangent equations, the y-intercepts (when $x=0$) are $y = 4$ and $y = -4$. The problem states that $q > 0$. Therefore, we must choose the tangent with the positive y-intercept. $q = 4$ The equation of the tangent T is: $y = \frac{4\sqrt{2}}{3}x + 4$ To find the x-intercept $p$, we set $y=0$ in the equation of T: $0 = \frac{4\sqrt{2}}{3}x + 4$ $\frac{4\sqrt{2}}{3}x = -4$ $x = -4 \times \frac{3}{4\sqrt{2}} = -\frac{3}{\sqrt{2}}$ Rationalizing the denominator: $p = x = -\frac{3\sqrt{2}}{2}$

Step 4: Calculate the value of pq

We have found $p = -\frac{3\sqrt{2}}{2}$ and $q = 4$. Now we compute their product: $pq = \left( -\frac{3\sqrt{2}}{2} \right) \times 4$ $pq = -3\sqrt{2} \times 2$ $pq = -6\sqrt{2}$ This corresponds to option A.