A triangle ABC has sides AB = 9 cm, BC = 12 cm, and AC = 15 cm. (a) Calculate the area of triangle ABC. (b) Calculate the size of angle BAC.

Eneo la pembetatu = $\frac{1}{2} \times \text{base} \times \text{height}$

The area of the triangle ABC is $\boxed{\text{54 cm}^2}$.

Tunajua $AB = 9\, \text{cm}$, $BC = 12\, \text{cm}$, na $AC = 15\, \text{cm}$.
Tunatumia $\sin(\angle BAC) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC}$.

The size of angle BAC is $\boxed{53.13^\circ}$.

Kipenyo $D = AC = 15\, \text{cm}$.

The radius of the circle is $\boxed{7.5\, \text{cm}}$.

Eneo la duara = $\pi r^2$.

The area of the shaded regions is $\boxed{122.71\, \text{cm}^2}$.

2) Two lines L1 and L2 intersect at a point M
L1: $2y - 3x - 6 = 0$
L2: $3y + x - 20 = 0$

(1) $2y - 3x = 6$
(2) $3y + x = 20$

Zidisha mlinganyo (2) kwa 3 ili kuondoa x.
$3 \times (3y + x) = 3 \times 20$
(3) $9y + 3x = 60$

$(2y - 3x) + (9y + 3x) = 6 + 60$
$11y = 66$

$3(6) + x = 20$
$18 + x = 20$
$x = 20 - 18$
$x = 2$
The coordinates of M are $\boxed{(2, 6)}$.