1. **Problem Statement:** Determine the values of $a$ that make the function $$f(x) = \begin{cases}\frac{\sqrt{3x-1} - \sqrt{2x}}{x-1}, & x \neq 1 \\ a, & x = 1 \end{cases}$$ continuous at $x=1$. 2. **Continuity Condition:** For $f(x)$ to be continuous at $x=1$, the limit of $f(x)$ as $x$ approaches 1 must equal $f(1)$, i.e., $$\lim_{x \to 1} f(x) = f(1) = a.$$ 3. **Calculate the limit:** We need to find $$\lim_{x \to 1} \frac{\sqrt{3x-1} - \sqrt{2x}}{x-1}.$$ Direct substitution gives $\frac{\sqrt{2} - \sqrt{2}}{0} = \frac{0}{0}$, an indeterminate form, so we use algebraic manipulation. 4. **Rationalize the numerator:** Multiply numerator and denominator by the conjugate: $$\frac{\sqrt{3x-1} - \sqrt{2x}}{x-1} \times \frac{\sqrt{3x-1} + \sqrt{2x}}{\sqrt{3x-1} + \sqrt{2x}} = \frac{(3x-1) - 2x}{(x-1)(\sqrt{3x-1} + \sqrt{2x})} = \frac{x-1}{(x-1)(\sqrt{3x-1} + \sqrt{2x})}.$$ 5. **Simplify:** Cancel $(x-1)$: $$\frac{1}{\sqrt{3x-1} + \sqrt{2x}}.$$ 6. **Evaluate the limit:** Substitute $x=1$: $$\frac{1}{\sqrt{3(1)-1} + \sqrt{2(1)}} = \frac{1}{\sqrt{2} + \sqrt{2}} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}.$$ 7. **Conclusion:** For continuity at $x=1$, set $$a = \frac{\sqrt{2}}{4}.$$ **Answer:** B) $\frac{\sqrt{2}}{4}$