Using the properties of detminants, evalulate. (i) `|{:(,23,6,11),(,36,5,26),(,63,13,37):}|` (ii) `|{:(,sqrt13+sqrt3,2sqrt5,sqrt5),(,sqrt15+sqrt26,5,sqrt10),(,3+sqrt65,sqrt15,5):}|`

Answer

To solve the given determinants, we will use properties of determinants step by step.

### Part (i)

We need to evaluate the determinant:

\[ D_1 = \begin{vmatrix} 23 & 6 & 11 \\ 36 & 5 & 26 \\ 63 & 13 & 37 \end{vmatrix} \]

**Step 1:** We can perform column operations to simplify the determinant. Let's modify the first column \(C_1\):

\[ C_1 \rightarrow C_1 - 2C_2 - C_3 \]

Calculating the new entries for \(C_1\):

- For the first row: \(23 - 2 \cdot 6 - 11 = 23 - 12 - 11 = 0\)
- For the second row: \(36 - 2 \cdot 5 - 26 = 36 - 10 - 26 = 0\)
- For the third row: \(63 - 2 \cdot 13 - 37 = 63 - 26 - 37 = 0\)

So, the modified determinant becomes:

\[ D_1 = \begin{vmatrix} 0 & 6 & 11 \\ 0 & 5 & 26 \\ 0 & 13 & 37 \end{vmatrix} \]

**Step 2:** Now, we see that the first column is entirely zero. According to the properties of determinants, if any row or column is entirely zero, the determinant is zero.

Thus, we have:

\[ D_1 = 0 \]

### Part (ii)

Now we evaluate the second determinant:

\[ D_2 = \begin{vmatrix} \sqrt{13} + \sqrt{3} & 2\sqrt{5} & \sqrt{5} \\ \sqrt{15} + \sqrt{26} & 5 & \sqrt{10} \\ 3 + \sqrt{65} & \sqrt{15} & 5 \end{vmatrix} \]

**Step 1:** We can also perform column operations here. Let's denote the columns as \(C_1\), \(C_2\), and \(C_3\). We can simplify by taking common factors out of the columns.

Taking \(\sqrt{3}\) from \(C_1\), \(\sqrt{5}\) from \(C_2\), and \(\sqrt{5}\) from \(C_3\):

\[ D_2 = \sqrt{3} \cdot \sqrt{5} \cdot \sqrt{5} \cdot \begin{vmatrix} 1 & 2\sqrt{5} & \sqrt{5} \\ \frac{\sqrt{15} + \sqrt{26}}{\sqrt{3}} & 5 & \frac{\sqrt{10}}{\sqrt{5}} \\ \frac{3 + \sqrt{65}}{\sqrt{3}} & \sqrt{15} & 5 \end{vmatrix} \]

**Step 2:** Now, we can simplify the determinant further. We can perform the operation \(C_2 \rightarrow C_2 - C_1\):

\[ D_2 = \sqrt{3} \cdot \sqrt{5} \cdot \sqrt{5} \cdot \begin{vmatrix} 1 & 2\sqrt{5} - 1 & \sqrt{5} \\ \frac{\sqrt{15} + \sqrt{26}}{\sqrt{3}} & 5 - \frac{\sqrt{15} + \sqrt{26}}{\sqrt{3}} & \frac{\sqrt{10}}{\sqrt{5}} \\ \frac{3 + \sqrt{65}}{\sqrt{3}} & \sqrt{15} - \frac{3 + \sqrt{65}}{\sqrt{3}} & 5 \end{vmatrix} \]

**Step 3:** After simplifying, we notice that two columns become identical, which implies that the determinant is zero.

Thus, we have:

\[ D_2 = 0 \]

### Final Answers

(i) \(D_1 = 0\)

(ii) \(D_2 = 0\)