Let $ABC$ be an equilateral triangle. $N$ is a point on the  side $AC$ such that $\vec{AC} = 7\vec{AN}$, $M$ is a point on the side  $AB$  such that $MN$ is parallel to $BC$ and $P$ is a point on the side $BC$  such that $MP$ is parallel to $AC$. Find the ratio of areas $\frac{ (MNP)}{(ABC)}$
1. **Define the Geometry and Given Conditions:**
   - Let $ABC$ be an equilateral triangle with side length $b$ and height $h$.
   - Point $N$ is on side $AC$ such that $\vec{AC} = 7\vec{AN}$, implying $AN = \frac{1}{7}AC = \frac{b}{7}$.
   - Point $M$ is on side $AB$ such that $MN \parallel BC$.
   - Point $P$ is on side $BC$ such that $MP \parallel AC$.

2. **Similarity of Triangles:**
   - Since $MN \parallel BC$, $\triangle AMN \sim \triangle ABC$ by AA similarity criterion.
   - Because $\triangle ABC$ is equilateral, $\triangle AMN$ is also equilateral.

3. **Calculate Lengths:**
   - Since $\triangle AMN \sim \triangle ABC$ and $AN = \frac{b}{7}$, the side length of $\triangle AMN$ is $MN = AN = \frac{b}{7}$.
   - The height of $\triangle AMN$ is $\frac{h}{7}$ because the height of an equilateral triangle is proportional to its side length.

4. **Area Calculation:**
   - The area of $\triangle ABC$ is given by:
     \[
     [ABC] = \frac{1}{2} \times b \times h
     \]
   - The area of $\triangle AMN$ is:
     \[
     [AMN] = \frac{1}{2} \times \frac{b}{7} \times \frac{h}{7} = \frac{1}{2} \times \frac{b}{7} \times \frac{h}{7} = \frac{bh}{98}
     \]

5. **Determine the Area of $\triangle MNP$:**
   - Since $MN \parallel BC$ and $MP \parallel AC$, $\triangle MNP$ is similar to $\triangle ABC$.
   - The ratio of the side lengths of $\triangle MNP$ to $\triangle ABC$ is $\frac{1}{7}$.
   - Therefore, the area ratio is the square of the side length ratio:
     \[
     \frac{[MNP]}{[ABC]} = \left(\frac{1}{7}\right)^2 = \frac{1}{49}
     \]

6. **Adjust for the Correct Ratio:**
   - The problem states that $MN = \frac{b}{7}$ and $MP \parallel AC$ implies that $MNP$ is a smaller equilateral triangle within $ABC$.
   - The correct ratio of areas is:
     \[
     \frac{[MNP]}{[ABC]} = \frac{\frac{1}{2} \times \frac{b}{7} \times \frac{h}{7}}{\frac{1}{2} \times b \times h} = \frac{\frac{bh}{98}}{\frac{bh}{2}} = \frac{1}{49}
     \]

The final answer is $\boxed{\frac{1}{49}}$