Let the position vector of the point $A$ be $\overrightarrow {OA}=x\hat i+y\hat j+z\hat k$

Given $|\overrightarrow{OA}|=10$

$\Rightarrow\:x^2+y^2+z^2=100$

Direction cosine $D.C.$ of $\overrightarrow {OA}=(cos 60^{\circ},cos 45^{\circ},n)$

$=(\large\frac{1}{2},\frac{1}{\sqrt 2},n)$......(i)

But $cos^2 60^{\circ}+cos^2 45^{\circ}+n^2=1$

$\Rightarrow\:\large\frac{1}{4}+\frac{1}{2}$$+n^2=1$

$\Rightarrow\:n^2=\large\frac{1}{4}$ or $n=\pm \large\frac{1}{2}$

$D.C$ of $\overrightarrow {OA}$ is also = $\bigg(\large\frac{x}{\sqrt {x^2+y^2+z^2}},\frac{y}{\sqrt {x^2+y^2+z^2}},\frac{z}{\sqrt {x^2+y^2+z^2}}\bigg)$

$=\big(\large\frac{x}{10},\frac{y}{10},\frac{z}{10}\big)$....(ii)

From (i) and (ii)

$\large\frac{x}{10}=\frac{1}{2},\:\frac{y}{10}=\frac{1}{\sqrt 2},$ and $\large\frac{z}{10}=\pm \frac{1}{2}$

$\Rightarrow\:x=5,y=5\sqrt 2,z=\pm 5$

$\Rightarrow\:\overrightarrow {OA}=5\hat i+5\sqrt 2\hat j\pm5\hat k$