Explanation

(d)
Let, $\overrightarrow{\mathbf{a}} = a_{1}\widehat{\dot{\mathbf{i}}} + a_{2}\widehat{\dot{\mathbf{j}}} + a_{3}\widehat{\mathbf{k}}$
$\because\ \ \overrightarrow{\mathbf{a}} \bullet \widehat{\dot{\mathbf{i}}} = 1 \Longrightarrow a_{1} = 1$
Since, $\overrightarrow{\mathbf{a}} \bullet \left( 2\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} \right) = 1$
$\Longrightarrow {2a}_{1} + a_{2} = 1$
$\Longrightarrow \ a_{2} = 1 - 2$
$\Longrightarrow \ a_{2} = - 1$
$and\ \overrightarrow{\mathbf{a}} \bullet \left( \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + 3\widehat{\mathbf{k}} \right) = 1$
$\Longrightarrow a_{1} + a_{2} + {3a}_{3} = 1$
$\Longrightarrow 1 - 1 + {3a}_{3} = 1$
$\Longrightarrow \ a_{3} = \frac{1}{3}$
$\therefore\overrightarrow{\mathbf{a}} = \widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} + \frac{1}{3}\widehat{\mathbf{k}} = \frac{1}{3}(3\widehat{\dot{\mathbf{i}}} - 3\widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}})\$