According to figure,

$\overrightarrow{\mathrm{OP}}=\overrightarrow{\mathrm{A}}, \overrightarrow{\mathrm{OQ}}=\overrightarrow{\mathrm{B}}$ and $\overrightarrow{\mathrm{QR}}=\overrightarrow{\mathrm{C}}$

Now $\vec{A} \cdot(\vec{B}+\vec{C})=($ Magnitude of $\vec{A}) \times($ Component of $\vec{B}+\vec{C}$ along the direction of $\vec{A})$

$=|\overrightarrow{\mathrm{A}}|(\mathrm{ON})$

$=|\overrightarrow{\mathrm{A}}|(\mathrm{OM}+\mathrm{MN})$

$=|\overrightarrow{\mathrm{A}}| \mathrm{OM}+|\overrightarrow{\mathrm{A}}| \mathrm{MN}$

$\therefore \quad \overrightarrow{\mathrm{A}} \cdot(\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}})=|\overrightarrow{\mathrm{A}}| \times($ component of $\overrightarrow{\mathrm{B}}$ along $\overrightarrow{\mathrm{A}})+|\overrightarrow{\mathrm{A}}|$ (component of $\overrightarrow{\mathrm{C}}$ along $\overrightarrow{\mathrm{A}}$ ) $\therefore \overrightarrow{\mathrm{A}} \cdot(\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}})=\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{C}}$