**Given:**
- In triangle ABC, l(AB)=6, l(BC)=7 and l(AC)=8
- AD is the perpendicular from A to BC
- AE is the angle bisector of angle A
- AF is the median of triangle ABC

**To find:**
- Details about the triangle based on the given information

**Solution:**

***Step 1: Finding the area of the triangle***

- The area of triangle ABC can be found using Heron's formula:

Area = sqrt(s(s-a)(s-b)(s-c))

where s is the semi-perimeter of the triangle and a, b, and c are the lengths of its sides.

- Semi-perimeter s = (l(AB) + l(BC) + l(AC))/2 = (6+7+8)/2 = 10.5

- Using the lengths of the sides, we get:

(s-a) = 4.5, (s-b) = 3.5, (s-c) = 2.5

- Plugging these values into the formula, we get:

Area = sqrt(10.5*4.5*3.5*2.5) = 10.5sqrt(15)

***Step 2: Finding the altitude AD***

- The altitude AD divides the triangle into two right triangles: ABD and ADC.
- Using Pythagoras theorem, we can find the length of AD:

AD^2 = AB^2 - BD^2

BD can be found using the fact that triangle AEB is similar to triangle ABC:

AE/AB = AE/(AE+EB) = AC/BC

Solving for EB, we get:

EB = AB * (AC-BC)/AC = 6 * 1/8 = 3/4

BD = AB - EB = 6 - 3/4 = 21/4

Plugging in the values, we get:

AD^2 = 6^2 - (21/4)^2

AD = 6sqrt(15)/8

***Step 3: Finding the length of AE***

- Let's assume that AE intersects BC at point X.
- Using the angle bisector theorem, we can find the length of BX:

BX/BC = AB/AC

BX = BC * AB/AC = 7 * 6/8 = 21/4

- Using the fact that triangle AEX is similar to triangle ABC, we can find the length of AE:

AE/AB = AX/AC

Solving for AE, we get:

AE = AB * AX/AC = 6 * 21/4 / 8 = 63/16

***Step 4: Finding the length of AF***

- Using the fact that AF is a median, we can find the length of AF:

AF^2 = 2(AB^2 + AC^2) - BC^2

Plugging in the values, we get:

AF^2 = 2(6^2 + 8^2) - 7^2 = 100

AF = 10

***Step 5: Summary of findings***

- Area

Area is 7