Explanation

(d)
(a) Let $\overrightarrow{\mathbf{p}} = x\widehat{\mathbf{i}} + y\widehat{\mathbf{j}}\mathcal{+ z}\widehat{\mathbf{k}}$ where at least one of $x,y,\ \mathcal{z}$ is non-zero. Let
$\overrightarrow{\mathbf{a}} = a_{1}\widehat{\mathbf{i}} + a_{2}\widehat{\mathbf{l}} + a_{3}\widehat{\mathbf{k}}$
$\overrightarrow{\mathbf{b}} = b_{1}\widehat{\mathbf{i}} + b_{2}\widehat{\mathbf{l}} + b_{3}\widehat{\mathbf{k}}$
$\overrightarrow{\mathbf{c}} = c_{1}\widehat{\mathbf{i}} + c_{2}\widehat{\mathbf{l}} + c_{3}\widehat{\mathbf{k}}\$
$\therefore$ By given conditions
$a_{1}x + a_{2}y + a_{3}\mathcal{z} = 0$
$b_{1}x + b_{2}y + b_{3}\mathcal{z} = 0$
$c_{1}x + c_{2}y + c_{3}\mathcal{z} = 0$
$\Rightarrow \ \left| \begin{matrix}
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3} \\
c_{1} & c_{3} & c_{3}
\end{matrix} \right| = 0$
$\Rightarrow \left\lbrack \ \overrightarrow{\mathbf{a}}\mathbf{\ }\overrightarrow{\mathbf{b}}\mathbf{\ }\overrightarrow{\mathbf{c}}\ \right\rbrack = 0$
$\Rightarrow \ \overrightarrow{\mathbf{a}}\mathbf{,}\overrightarrow{\mathbf{b}}\mathbf{,}\overrightarrow{\mathbf{c}}$ are coplanar.
(b) Vectors are coplanar, if
$\left| \begin{matrix}
1 & 3 & 0 \\
2 & 0 & 1 \\
0 & 1 & 1
\end{matrix} \right| = 0$
$ie,\ - 7 = 0$
Which is not possible.
( c ) $\overrightarrow{\mathbf{a}} \times \left( \ \overrightarrow{\mathbf{b}}\mathbf{\times}\overrightarrow{\mathbf{c}}\ \right) = \left( \mathbf{\ }\overrightarrow{\mathbf{a}}\mathbf{\bullet}\overrightarrow{\mathbf{c}}\mathbf{\ } \right)\overrightarrow{\mathbf{b}} - \left( \mathbf{\ }\overrightarrow{\mathbf{a}}\mathbf{\bullet}\overrightarrow{\mathbf{b}}\mathbf{\ } \right)\overrightarrow{\mathbf{c}}$
$\Rightarrow \ \overrightarrow{\mathbf{a}} \times \left( \ \overrightarrow{\mathbf{b}}\mathbf{\times}\overrightarrow{\mathbf{c}}\mathbf{\ } \right)$ is coplanar with $\overrightarrow{b}\ and\ \overrightarrow{c}$.
(d) $\left| \overrightarrow{\mathbf{a}} \right|\mathbf{=}\left| \overrightarrow{\mathbf{b}} \right| = 1$
$\therefore\left| \overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}} \right|^{2} = \left( \overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}} \right)\mathbf{\bullet (}\overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}})$
$= \left| \overrightarrow{\mathbf{a}} \right|^{2} + \left| \overrightarrow{\mathbf{b}} \right|^{2} + 2\overrightarrow{\mathbf{a}}\mathbf{\bullet}\overrightarrow{\mathbf{b}}$
$= 1 + 1 = 2 \bullet 1 \bullet 1\cos{\frac{\pi}{3} = 3}$
$\Rightarrow \ \left| \overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}} \right| = \sqrt{3} > 1$