The co-ordinates of the point from which the length of tangents to the circles
`3x^(2)+3y^(2)+4x-6y-1=0`,
`2x^(2)+2y^(2)-3x-2y-4=0`
and `2x^(2)+2y^(2)-x+y-1=0` be equal is

Options

A. `((10)/(21), (14)/(63))`
B. (1, 0)
C. `(-(10)/(21), -(31)/(63))`
D. none

Answer

To find the coordinates of the point from which the lengths of tangents to the given circles are equal, we will follow these steps:

### Step 1: Convert the equations of the circles to standard form
The equations of the circles are given as:
1. \(3x^2 + 3y^2 + 4x - 6y - 1 = 0\)
2. \(2x^2 + 2y^2 - 3x - 2y - 4 = 0\)
3. \(2x^2 + 2y^2 - x + y - 1 = 0\)

We will divide each equation by the coefficient of \(x^2\) and \(y^2\) to convert them into standard form.

1. For the first circle:
\[ x^2 + y^2 + \frac{4}{3}x - 2y - \frac{1}{3} = 0 \]

2. For the second circle:
\[ x^2 + y^2 - \frac{3}{2}x - y - 2 = 0 \]

3. For the third circle:
\[ x^2 + y^2 - \frac{1}{2}x + \frac{1}{2}y - \frac{1}{2} = 0 \]

### Step 2: Find the radical axes
The radical axis of two circles can be found by subtracting their equations.

1. For circles 1 and 2:
\[ (3x^2 + 3y^2 + 4x - 6y - 1) - (2x^2 + 2y^2 - 3x - 2y - 4) = 0 \]
Simplifying gives:
\[ x^2 + y^2 + \frac{17}{6}x - 4y + \frac{5}{3} = 0 \]

2. For circles 2 and 3:
\[ (2x^2 + 2y^2 - 3x - 2y - 4) - (2x^2 + 2y^2 - x + y - 1) = 0 \]
Simplifying gives:
\[ -x - \frac{3}{2} = 0 \quad \Rightarrow \quad x = -\frac{3}{2} \]

3. For circles 1 and 3:
\[ (3x^2 + 3y^2 + 4x - 6y - 1) - (2x^2 + 2y^2 - x + y - 1) = 0 \]
Simplifying gives:
\[ x + \frac{11}{6}x + \frac{5}{2}y - \frac{1}{6} = 0 \]

### Step 3: Solve the equations of the radical axes
Now we will solve the equations obtained from the radical axes to find the radical center.

1. From the first radical axis equation, we can express \(y\) in terms of \(x\) and substitute into the second radical axis equation.

2. Solving these equations will yield the coordinates of the radical center, which will be the point from which the lengths of the tangents to all three circles are equal.

### Step 4: Calculate the coordinates
After solving the equations, we find:
\[ x = -\frac{16}{21}, \quad y = \frac{31}{63} \]

### Final Answer
Thus, the coordinates of the point from which the lengths of tangents to the circles are equal are:
\[ \left(-\frac{16}{21}, \frac{31}{63}\right) \]