$O$ is the origin and $A$ is $(a, b, c)$. Find the direction cosines of the line $OA$ and the equation of the plane passing through $A$ at a right angle to $OA$.

The coordinates of the origin $O$ are $(0, 0, 0)$ and point $A$ are $(a, b, c)$.
The direction ratios of the line $OA$ are $(a-0, b-0, c-0) = (a, b, c)$.
The length of $OA$ is $\sqrt{a^2 + b^2 + c^2}$.
Therefore,the direction cosines of the line $OA$ are $\frac{a}{\sqrt{a^2 + b^2 + c^2}}, \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \frac{c}{\sqrt{a^2 + b^2 + c^2}}$.
The normal vector $\vec{n}$ to the plane is $\overrightarrow{OA} = a\hat{i} + b\hat{j} + c\hat{k}$.
The equation of a plane passing through a point $\vec{a} = (a, b, c)$ with normal vector $\vec{n}$ is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$.
Substituting $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$,$\vec{a} = a\hat{i} + b\hat{j} + c\hat{k}$,and $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$:
$(x\hat{i} + y\hat{j} + z\hat{k} - (a\hat{i} + b\hat{j} + c\hat{k})) \cdot (a\hat{i} + b\hat{j} + c\hat{k}) = 0$
$(x-a)a + (y-b)b + (z-c)c = 0$
$ax - a^2 + by - b^2 + cz - c^2 = 0$
$ax + by + cz = a^2 + b^2 + c^2$.