Given, $$z = {{(1 - t){z_1} + t\,{z_2}} \over {(1 - t) + t}}$$
Clearly, z divides z1 and z2 in the ratio of t : (1 $$-$$ t), 0 < t < 1
$$\Rightarrow$$ AP + BP = AB
i.e., $$|z - {z_1}| + |z - {z_2}| = |{z_1} - {z_2}|$$
$$\Rightarrow$$ Option (a) is true.
and $$\arg (z - {z_1}) = \arg ({z_2} - z) = \arg ({z_2} - {z_1})$$
$$\Rightarrow$$ (b) is false and (d) is true.
Also, $$\arg (z - {z_1}) = \arg ({z_2} - {z_1})$$
$$ \Rightarrow \arg \left( {{{z - {z_1}} \over {{z_2} - {z_1}}}} \right) = 0$$
$$\therefore$$ $${{z - {z_1}} \over {{z_2} - {z_1}}}$$ is purely real.
$$ \Rightarrow {{z - {z_1}} \over {{z_2} - {z_1}}} = {{\overline z - {{\overline z }_1}} \over {{{\overline z }_2} - {{\overline z }_1}}}$$
or, $$\left| {\matrix{ {z - {z_1}} & {\overline z - {{\overline z }_1}} \cr {{z_2} - {z_1}} & {{{\overline z }_2} - {{\overline z }_1}} \cr } } \right| = 0$$
$$\therefore$$ Option (c) is correct.
Hence, (a, c, d) is the correct option.
