In a triangle ABC, AB = 5 , BC = 7, AC = 6. A point P is in the plane such that it is at distance '2' units from AB and 3 units form AC then its distance from BC

Options

A. is `(12 sqrt6-28)/(7)` when P is inside the trinagle
B. may be `(12 sqrt6 -8)/(7)` when P is outside the triangle
C. may be `(12 sqrt6 + 14)/(7)` when P is inside the triangle
D. may be `(12 sqrt6 + 14)/(7)` when P is outside the triangle

Explanation

To solve the problem, we need to find the distance from point P to side BC of triangle ABC, given that point P is 2 units away from side AB and 3 units away from side AC.

### Step-by-Step Solution:

1. **Identify the Triangle and Sides**:
- We have triangle ABC with sides:
- \( AB = 5 \)
- \( BC = 7 \)
- \( AC = 6 \)

2. **Calculate the Semi-perimeter (s)**:
- The semi-perimeter \( s \) of triangle ABC is given by:
\[ s = \frac{AB + BC + AC}{2} = \frac{5 + 7 + 6}{2} = 9 \]

3. **Calculate the Area (A) of Triangle ABC using Heron's Formula**:
- Heron's formula states:
\[ A = \sqrt{s(s-a)(s-b)(s-c)} \]
- Here, let:
- \( a = BC = 7 \)
- \( b = AC = 6 \)
- \( c = AB = 5 \)
- Substituting the values:
\[ A = \sqrt{9(9-7)(9-6)(9-5)} = \sqrt{9 \times 2 \times 3 \times 4} = \sqrt{216} = 6\sqrt{6} \]

4. **Set Up the Area Equation with Point P**:
- The area of triangle ABC can also be expressed in terms of the distances from point P to the sides:
\[ A = \frac{1}{2} \times AB \times d_{AB} + \frac{1}{2} \times AC \times d_{AC} + \frac{1}{2} \times BC \times d_{BC} \]
- Here, \( d_{AB} = 2 \) (distance from P to AB), \( d_{AC} = 3 \) (distance from P to AC), and \( d_{BC} = x \) (distance from P to BC).
- Thus, we can write:
\[ 6\sqrt{6} = \frac{1}{2} \times 5 \times 2 + \frac{1}{2} \times 6 \times 3 + \frac{1}{2} \times 7 \times x \]

5. **Simplify the Equation**:
- Calculate the known areas:
\[ 6\sqrt{6} = 5 + 9 + \frac{7x}{2} \]
- Combine the constants:
\[ 6\sqrt{6} = 14 + \frac{7x}{2} \]

6. **Isolate x**:
- Rearranging gives:
\[ 6\sqrt{6} - 14 = \frac{7x}{2} \]
- Multiply both sides by 2:
\[ 12\sqrt{6} - 28 = 7x \]
- Finally, solve for \( x \):
\[ x = \frac{12\sqrt{6} - 28}{7} \]

### Final Answer:
The distance from point P to side BC is:
\[ x = \frac{12\sqrt{6} - 28}{7} \]