6. Given the system of equations $\left\{\begin{array}{l}|3 x|+|4 y|+|48-3 x-4 y|=48, \\ (x-8)^{2}+\left(y+6 \cos \frac{a \pi}{2}\right)^{2}=(a+4)^{2} \text {. }\end{array}\right.$

a) Sketch on the $(x ; y)$ plane the set of points satisfying the first equation of the system, and find the area of the resulting figure.

b) Find all values of the parameter $a$ for which the system has exactly three solutions.
Answer: a) 96; b) $a=6, a=-14$.

Solution. Note that the equality $|a|+|b|+|c|=a+b+c$ holds if and only if the numbers $a, b$, and $c$ are non-negative (since if at least one of them is negative, the left side is greater than the right side). Therefore, the first equation is equivalent to the system of inequalities

$$
\left\{\begin{array} { l } 
{ 3 x \geq 0 , } \\
{ 4 y \geq 0 , } \\
{ 4 8 - 3 x - 4 y \geq 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x \geq 0 \\
y \geq 0 \\
3 x+4 y \leq 48
\end{array}\right.\right.
$$

This system defines a triangle on the plane with vertices $E(16 ; 0), G(0 ; 12), N(0 ; 0)$, the area of which is 96. The second equation of the original system defines a circle (or a point when $a=-4$). The system can have exactly three solutions only in one case: when the circle defined by the second equation is circumscribed around the triangle $E G N$. The center of the circle circumscribed around a right triangle is the midpoint of the hypotenuse, and its radius is half the hypotenuse, from which we obtain the conditions

$$
-6 \cos \frac{a \pi}{2}=6,(a+4)^{2}=10^{2}
$$

The second equation gives two values of the parameter: $a=-14$ and $a=6$. By substitution, we verify that both satisfy the first equation.