9. Without using a calculator, find the value of \( p \) such that \( \left(\frac{2}{\sqrt{75}}-\frac{\sqrt{147}}{3}\right) \div \sqrt{2}=p \sqrt{6} \).

To solve for \( p \) in the equation \[ \left(\frac{2}{\sqrt{75}} - \frac{\sqrt{147}}{3}\right) \div \sqrt{2} = p \sqrt{6}, \] we first simplify the left-hand side of the equation. 1. Simplifying \( \sqrt{75} \): \[ \sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}. \] Thus, \[ \frac{2}{\sqrt{75}} = \frac{2}{5\sqrt{3}}. \] 2. Simplifying \( \sqrt{147} \): \[ \sqrt{147} = \sqrt{49 \cdot 3} = 7\sqrt{3}. \] Therefore, \[ \frac{\sqrt{147}}{3} = \frac{7\sqrt{3}}{3} = \frac{7}{3} \sqrt{3}. \] 3. Now substituting these simplifications into the expression: \[ \frac{2}{5\sqrt{3}} - \frac{7\sqrt{3}}{3}. \] To subtract these fractions, we need a common denominator. The least common multiple of \( 5\sqrt{3} \) and \( 3 \) is \( 15\sqrt{3} \). Rewriting the first term: \[ \frac{2}{5\sqrt{3}} = \frac{2 \cdot 3}{5 \cdot 3\sqrt{3}} = \frac{6}{15\sqrt{3}}. \] Rewriting the second term: \[ \frac{7\sqrt{3}}{3} = \frac{7\sqrt{3} \cdot 5\sqrt{3}}{3 \cdot 5\sqrt{3}} = \frac{35(\sqrt{3})^2}{15\sqrt{3}} = \frac{105}{15\sqrt{3}}. \] 4. Now we can perform the subtraction: \[ \frac{6}{15\sqrt{3}} - \frac{35}{15\sqrt{3}} = \frac{6 - 35}{15\sqrt{3}} = \frac{-29}{15\sqrt{3}}. \] 5. Next, divide this result by \( \sqrt{2} \): \[ \left(\frac{-29}{15\sqrt{3}}\right) \div \sqrt{2} = \frac{-29}{15\sqrt{3}} \cdot \frac{1}{\sqrt{2}} = \frac{-29}{15\sqrt{6}}. \] 6. Now, we express this as \[ p\sqrt{6}. \] To make the left-hand side in the same form, multiply the numerator and the denominator by \( \sqrt{6} \): \[ \frac{-29\sqrt{6}}{15\cdot 6} = \frac{-29\sqrt{6}}{90}. \] 7. Setting this equal to \( p\sqrt{6} \), \[ p\sqrt{6} = \frac{-29\sqrt{6}}{90}. \] This implies that \[ p = \frac{-29}{90}. \] Thus, the value of \( p \) is \[ \boxed{\frac{-29}{90}}. \]