Given the three circles
` x^(2) + y^(2) - 16 x + 60 = 0`
` 3 x^2 + 3y ^(2) - 36 x + 81 = 0`
and ` x^(2) + y^(2) - 16 x - 12 y + 84 = 0`
find (1) the point from which the tangents to them are equal in length, and (2) this length.

Answer

To solve the problem, we need to find the point from which the tangents to the three given circles are equal in length, and then we will calculate that length.

### Step 1: Rewrite the equations of the circles in standard form

1. **Circle 1:**
\[ x^2 + y^2 - 16x + 60 = 0 \implies (x - 8)^2 + y^2 = 4 \]
This circle has center \( (8, 0) \) and radius \( 2 \).

2. **Circle 2:**
\[ 3x^2 + 3y^2 - 36x + 81 = 0 \implies x^2 + y^2 - 12x + 27 = 0 \implies (x - 6)^2 + (y - 0)^2 = 9 \]
This circle has center \( (6, 0) \) and radius \( 3 \).

3. **Circle 3:**
\[ x^2 + y^2 - 16x - 12y + 84 = 0 \implies (x - 8)^2 + (y - 6)^2 = 4 \]
This circle has center \( (8, 6) \) and radius \( 2 \).

### Step 2: Set up the equations for tangent lengths

Let the point from which the tangents are drawn be \( P(\alpha, \beta) \). The length of the tangent from a point \( P(x_1, y_1) \) to a circle \( (x - h)^2 + (y - k)^2 = r^2 \) is given by:
\[ L^2 = (x_1 - h)^2 + (y_1 - k)^2 - r^2 \]

Using this formula, we can find the lengths of the tangents from point \( P(\alpha, \beta) \) to each of the circles:

1. **For Circle 1:**
\[ L_1^2 = (\alpha - 8)^2 + (\beta - 0)^2 - 2^2 = (\alpha - 8)^2 + \beta^2 - 4 \]

2. **For Circle 2:**
\[ L_2^2 = (\alpha - 6)^2 + (\beta - 0)^2 - 3^2 = (\alpha - 6)^2 + \beta^2 - 9 \]

3. **For Circle 3:**
\[ L_3^2 = (\alpha - 8)^2 + (\beta - 6)^2 - 2^2 = (\alpha - 8)^2 + (\beta - 6)^2 - 4 \]

### Step 3: Set up the equations for equal lengths

We need to find \( \alpha \) and \( \beta \) such that \( L_1^2 = L_2^2 = L_3^2 \).

1. **Setting \( L_1^2 = L_2^2 \):**
\[ (\alpha - 8)^2 + \beta^2 - 4 = (\alpha - 6)^2 + \beta^2 - 9 \]
Simplifying gives:
\[ (\alpha - 8)^2 - (\alpha - 6)^2 + 5 = 0 \]
Expanding and simplifying:
\[ -4\alpha + 32 + 5 = 0 \implies -4\alpha + 37 = 0 \implies \alpha = \frac{37}{4} \]

2. **Setting \( L_2^2 = L_3^2 \):**
\[ (\alpha - 6)^2 + \beta^2 - 9 = (\alpha - 8)^2 + (\beta - 6)^2 - 4 \]
Simplifying gives:
\[ (\alpha - 6)^2 - (\alpha - 8)^2 + \beta^2 - (\beta - 6)^2 - 5 = 0 \]
Expanding and simplifying:
\[ 4\alpha - 32 + 6\beta - 36 - 5 = 0 \implies 4\alpha + 6\beta - 73 = 0 \]
Substituting \( \alpha = \frac{37}{4} \):
\[ 4 \cdot \frac{37}{4} + 6\beta - 73 = 0 \implies 37 + 6\beta - 73 = 0 \implies 6\beta = 36 \implies \beta = 6 \]

### Step 4: Find the length of the tangents

Now that we have \( \alpha = \frac{37}{4} \) and \( \beta = 6 \), we can find the length of the tangents using any of the tangent length formulas. Let's use \( L_1^2 \):

\[ L_1^2 = \left(\frac{37}{4} - 8\right)^2 + (6 - 0)^2 - 4 \]
Calculating:
\[ L_1^2 = \left(\frac{37}{4} - \frac{32}{4}\right)^2 + 6^2 - 4 = \left(\frac{5}{4}\right)^2 + 36 - 4 = \frac{25}{16} + 32 = \frac{25 + 512}{16} = \frac{537}{16} \]
Thus, the length \( L_1 = \sqrt{\frac{537}{16}} = \frac{\sqrt{537}}{4} \).

### Final Answers

1. The point from which the tangents to the circles are equal in length is \( \left(\frac{37}{4}, 6\right) \).
2. The length of the tangents is \( \frac{\sqrt{537}}{4} \).