`(i)` Find the value of `'a'` if the lines `3x-2y+8=0`, `2x+y+3=0` and `ax+3y+11=0` are concurrent.
`(ii)` If the lines `y=m_(1)x+c_(1)`, `y=m_(2)x+c_(2)` and `y=m_(3)x+c_(3)` meet at point then shown that :
`c_(1)(m_(2)-m_(3))+c_(2)(m_(3)-m_(1))+c_(3)(m_(1)-m_(2))=0`

Answer

To solve the given problem step by step, we will break it down into two parts as per the question.

### Part (i): Finding the value of 'a'

We need to find the value of 'a' such that the lines:
1. \(3x - 2y + 8 = 0\)
2. \(2x + y + 3 = 0\)
3. \(ax + 3y + 11 = 0\)

are concurrent.

**Step 1: Write the equations in standard form.**
The equations are already in the standard form \(a_1x + b_1y + c_1 = 0\), \(a_2x + b_2y + c_2 = 0\), and \(a_3x + b_3y + c_3 = 0\).

- For the first line: \(a_1 = 3\), \(b_1 = -2\), \(c_1 = 8\)
- For the second line: \(a_2 = 2\), \(b_2 = 1\), \(c_2 = 3\)
- For the third line: \(a_3 = a\), \(b_3 = 3\), \(c_3 = 11\)

**Step 2: Set up the determinant for concurrency.**
The lines are concurrent if the determinant of the coefficients is zero:

\[ \begin{vmatrix} 3 & -2 & 8 \\ 2 & 1 & 3 \\ a & 3 & 11 \end{vmatrix} = 0 \]

**Step 3: Calculate the determinant.**
Calculating the determinant:

\[ = 3 \begin{vmatrix} 1 & 3 \\ 3 & 11 \end{vmatrix} - (-2) \begin{vmatrix} 2 & 3 \\ a & 11 \end{vmatrix} + 8 \begin{vmatrix} 2 & 1 \\ a & 3 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \(\begin{vmatrix} 1 & 3 \\ 3 & 11 \end{vmatrix} = (1)(11) - (3)(3) = 11 - 9 = 2\)
2. \(\begin{vmatrix} 2 & 3 \\ a & 11 \end{vmatrix} = (2)(11) - (3)(a) = 22 - 3a\)
3. \(\begin{vmatrix} 2 & 1 \\ a & 3 \end{vmatrix} = (2)(3) - (1)(a) = 6 - a\)

Substituting back into the determinant:

\[ = 3(2) + 2(22 - 3a) + 8(6 - a) \]
\[ = 6 + 44 - 6a + 48 - 8a \]
\[ = 98 - 14a \]

**Step 4: Set the determinant to zero.**
Setting the determinant equal to zero for concurrency:

\[ 98 - 14a = 0 \]

**Step 5: Solve for 'a'.**
\[ 14a = 98 \implies a = \frac{98}{14} = 7 \]

Thus, the value of \(a\) is **7**.