6. In $\triangle \mathrm{ABC}$, $\angle \mathrm{A}=\arccos \frac{7}{8}, \mathrm{BC}=\mathrm{a}$, the altitude from vertex $\mathrm{A}$ is equal to the sum of the other two altitudes. Find the area of $\triangle \mathrm{ABC}$.
6. Let $\mathrm{AB}=\mathrm{x}, \mathrm{AC}=\mathrm{y}$ (Figure 6).

At this time, the area of $\triangle \mathrm{ABC}$ is
$$
S=\frac{1}{2} x y \sin B A C=x y \frac{\sqrt{15}}{16}
$$

The altitudes of the triangle can be expressed as:
$$
\mathrm{h}_{\mathrm{a}}=\frac{2 \mathrm{~S}}{\mathrm{a}}, \mathrm{h}_{\mathrm{x}}=\frac{2 \mathrm{~S}}{\mathrm{x}}, \mathrm{h}_{\mathrm{y}}=\frac{2 \mathrm{~S}}{\mathrm{y}} \text {. }
$$

According to the problem's conditions, if $\mathrm{h}_{2}=\mathrm{i} \mathbf{x}+1, \ldots$. $(2)$, we get
$$
x+y=\frac{6 y}{a}
$$

Using the cosine rule, we can obtain the expression for side $\mathrm{BC}$ in terms of $x$ and $y$:
$$
x^{2}+y^{2}-\frac{7}{4} x y=a^{2} \text {. }
$$

Solving the system of equations (3) and (4). Squaring both sides of equation (3) and then subtracting equation (4), we get
$$
\frac{(x y)^{2}}{a^{2}}-\frac{15}{4} x y-a^{2}=0 .
$$

From this equation, we get $x y=4 a^{2}$ (discarding the negative root). Thus, from equation (1), we can obtain
$$
S=a^{2} \sqrt{15}
$$