Using step by step method to solve all the questions

The UDL is $10 \text{ kN/m}$ over a length of $3 \text{ m}$.
The equivalent point load $F_{UDL}$ is:

$$F_{UDL} = 10 \text{ kN/m} \times 3 \text{ m} = 30 \text{ kN}$$

This equivalent load acts at the centroid of the UDL, which is $3 \text{ m} / 2 = 1.5 \text{ m}$ from support A.

Assume forces acting to the right are positive.

$$\sum F_x = 0$$ $$R_{Ax} + 15 \text{ kN} = 0$$ $$R_{Ax} = -15 \text{ kN}$$

The negative sign indicates that $R_{Ax}$ acts to the left.

$$\boxed{R_{Ax} = 15 \text{ kN (to the left)}}$$

Assume counter-clockwise moments are positive. The horizontal force of $15 \text{ kN}$ passes through point A's horizontal line of action, so it creates no moment about A.

$$\sum M_A = 0$$ $$-(F_{UDL} \times 1.5 \text{ m}) - (20 \text{ kN} \times 5 \text{ m}) - (10 \text{ kN} \times 7 \text{ m}) + (R_B \times 7 \text{ m}) = 0$$ $$-(30 \text{ kN} \times 1.5 \text{ m}) - (20 \text{ kN} \times 5 \text{ m}) - (10 \text{ kN} \times 7 \text{ m}) + (R_B \times 7 \text{ m}) = 0$$ $$-45 \text{ kNm} - 100 \text{ kNm} - 70 \text{ kNm} + 7 R_B \text{ m} = 0$$ $$-215 \text{ kNm} + 7 R_B \text{ m} = 0$$ $$7 R_B = 215 \text{ kN}$$ $$R_B = \frac{215}{7} \text{ kN}$$ $$\boxed{R_B \approx 30.71 \text{ kN (upward)}}$$

Assume upward forces are positive.

$$\sum F_y = 0$$ $$R_{Ay} + R_B - F_{UDL} - 20 \text{ kN} - 10 \text{ kN} = 0$$ $$R_{Ay} + \frac{215}{7} \text{ kN} - 30 \text{ kN} - 20 \text{ kN} - 10 \text{ kN} = 0$$ $$R_{Ay} + \frac{215}{7} \text{ kN} - 60 \text{ kN} = 0$$ $$R_{Ay} = 60 \text{ kN} - \frac{215}{7} \text{ kN}$$ $$R_{Ay} = \frac{60 \times 7 - 215}{7} \text{ kN}$$ $$R_{Ay} = \frac{420 - 215}{7} \text{ kN}$$ $$R_{Ay} = \frac{205}{7} \text{ kN}$$ $$\boxed{R_{Ay} \approx 29.29 \text{ kN (upward)}}$$