If a unit vector $\overrightarrow{r}$ makes an angle $\dfrac{\pi }{3}$ with $\overset{\hat{\ }}{\mathop{i}}\,$ , $\dfrac{\pi }{4}$ with $\widehat{j}$ and $\theta \in \left( 0,\pi \right)$ with $\widehat{k}$,then a value of $\theta $ is
(a) $\dfrac{5\pi }{12}$
(b) $\dfrac{5\pi }{6}$
(c) $\dfrac{2\pi }{3}$
(d) $\dfrac{\pi }{3}$

Hint:We know that cosines of the angles made by a unit vector with the three coordinate axes are known as directional cosine of that vector. Suppose if a unit vector makes an angle $\alpha ,\beta \,\,\text{and }\gamma $ with x, y and z axes respectively, then we have ${{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1$ . So using this property, in this problem we will find the unknown angle.

Complete step by step answer:
Since , $\overset{\hat{\ }}{\mathop{i}}\,$ , $\widehat{j}$, $\widehat{k}$ gives the direction of x, y and z axes respectively.
Here in this problem we have, angle made by unit vector $\overrightarrow{r}$ with x-axis is $\dfrac{\pi }{3}$, with y-axzis is $\dfrac{\pi }{4}$ and with z-axis is $\theta $, then using the property of directional cosine that is if a unit vector makes an angle $\alpha ,\beta \,\,\text{and }\gamma $ with x, y and z axes respectively, then we have ${{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1$, so here we have $\alpha =\dfrac{\pi }{3}$ , $\beta =\dfrac{\pi }{4}$ and $\gamma =\theta $
$\begin{align}
  & \therefore {{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1 \\
 & \Rightarrow {{\cos }^{2}}\dfrac{\pi }{3}+{{\cos }^{2}}\dfrac{\pi }{4}+{{\cos }^{2}}\theta =1 \\
\end{align}$
Simplifying further we have,
$\begin{align}
  & {{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}+{{\cos }^{2}}\theta =1 \\
 & \Rightarrow \dfrac{1}{4}+\dfrac{1}{2}+{{\cos }^{2}}\theta =1 \\
\end{align}$
Bringing all like terms to same side, we have
$\begin{align}
  & {{\cos }^{2}}\theta =1-\left( \dfrac{1}{4}+\dfrac{1}{2} \right) \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,=1-\dfrac{3}{4} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{1}{4} \\
\end{align}$
Taking square root both sides, we have
$\begin{align}
  & \cos \theta =\sqrt{\dfrac{1}{4}} \\
 & \,\,\,\,\,\,\,\,\,\,\,=\dfrac{1}{2} \\
\end{align}$
Now ,
 $\begin{align}
  & \theta ={{\cos }^{-1}}\dfrac{1}{2} \\
 & \,\,\,\,=\dfrac{\pi }{3}\text{ or 2}\pi -\dfrac{\pi }{3} \\
 & \,\,\,\,=\dfrac{\pi }{3}\text{ or }\dfrac{5\pi }{3} \\
\end{align}$
But , $\theta \in \left( 0,\pi \right)$ so $\theta =\dfrac{\pi }{3}$
Hence, the correct answer is option (d)

Note:
Always remember directional cosine is cosines of the angle made by the vector, with the three axes. Also, angle made by a vector with the three axes is always less than or equal to 180 degrees. Here, don’t take the values like $\theta =\dfrac{5\pi }{3}$ as $\theta $ is angle made by the unit vector with z-axis and also it is given that $\theta \in \left( 0,\pi \right)$. So be careful while taking the value of $\theta $ here.