Question:

Step 1: Analyze the given conditions for each row in the table to understand what is being asked. We have four rows (I) to (IV), each representing a different quadratic equation with specific conditions. We need to match these conditions with the number of non-positive integral values of 'k' and the number of prime values of 'k' given in columns 2 and 3, respectively. Step 2: For row (I), the quadratic equation $x^2 - 8x + k^2 - 6k = 0$ has real roots $\alpha$ and $\beta$ such that $\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = 2$. This condition leads to the equation $\dfrac{\alpha^2 + \beta^2}{\alpha\beta} = 2$. Substituting $\alpha + \beta = 8$ and $\alpha\beta = k^2 - 6k$ into the equation gives $64 = 4(k^2-6k)$, which simplifies to $k^2 - 6k - 16 = 0$. Solving this quadratic equation yields $k = 8$ or $k = -2$. The non-positive integral value of $k$ is $-2$, giving a count of 1, and there are no prime values of $k$ here, giving a count of 0. Step 3: For row (II), the quadratic equation $(k-2)x^2 - (8-2k)x + (3k+8) = 0$ has one negative and one positive root. This implies that the product of the roots, $\dfrac{3k+8}{k-2}$, is negative. The inequality $-\dfrac{8}{3} < k < 2$ must hold for the roots to have opposite signs. The non-positive integral values of $k$ in this range are $-2, -1, 0$, giving a count of 3, and there are no prime values of $k$ in this range, giving a count of 0. Step 4: For row (III), the quadratic equation $4x^2 - 2kx + 1 = 0$ has real roots with a difference less than $\sqrt{3}$. The roots are given by $x = \dfrac{2k \pm \sqrt{4k^2-16}}{8}$. The difference between the roots is $\dfrac{\sqrt{k^2-4}}{2} < \sqrt{3}$, leading to $k^2 < 16$, or $-4 < k < 4$. For real roots, $k \leq -2$ or $k \geq 2$. Combining these conditions gives $k \in (-4, -2] \cup [2, 4)$. The non-positive integral values of $k$ are $-3$ and $-2$, giving a count of 2, and the prime values of $k$ are 2 and 3, giving a count of 2. Step 5: For row (IV), the quadratic expression $2kx^2 - (4k-5)x - 10 < 0$ is negative for exactly three distinct integral values of $x$. For the expression to be negative for finitely many values, $k > 0$ is required. The roots of the quadratic equation $2kx^2 - (4k-5)x - 10 = 0$ have opposite signs since their product is negative. Let the roots be $\alpha < 0 < \beta$. The expression is negative for $x \in (\alpha, \beta)$. For exactly 3 integral values, we need exactly 3 integers in $(\alpha, \beta)$. Testing $k = 2$ yields roots $x = 2$ and $x = -1.25$, and the integers in $(-1.25, 2)$ are $-1, 0, 1$, which are three integers. Thus, $k = 2$ satisfies the condition. The non-positive integral value count is 0 since $k > 0$, and the prime value count is 1 since $k = 2$ is prime. Step 6: Now, let's match the conditions for each row with the given options. Row (I) matches (ii)(P), row (II) matches (iv)(P), row (III) matches (iii)(R), and row (IV) matches (i)(Q). The only correct combination among the given options is the one that matches row (IV) with (i)(Q), which corresponds to option (d). Therefore, the correct answer is option (d), which is the combination (IV)(i)(Q). The final answer is $\boxed{3}$.