From the figure, the four diagonals are $~\vec{OP},~\vec{AR},~\vec{BS},~\vec{CQ}~$.
Direction ratios of $~\vec{OP}:a−o,a−o,a−o=a,a,a=1,1,1~$
Direction ratios of $~\vec{AR}:o−a,a−o,a−o=−a,a,a=−1,1,1~$
Direction ratios of $~\vec{BS}:a−o,o−a,a−o=a,−a,a=1,−1,1~$
Direction ratios of $~\vec{CQ}:a−o,a−o,o−a=a,a,−a=1,1,−1~$
$∴$ Direction cosine (DC)'s of $~\vec{OP}~$ are $~\dfrac{1}{\sqrt 3}~,\dfrac{1}{\sqrt 3},~\dfrac{1}{\sqrt 3}~$
DC's of $~\vec{AR}~$ are $~-\dfrac{1}{\sqrt 3}~,\dfrac{1}{\sqrt 3},~\dfrac{1}{\sqrt 3}~$
DC's of $~\vec{BS}~$ are $~\dfrac{1}{\sqrt 3}~,-\dfrac{1}{\sqrt 3},~\dfrac{1}{\sqrt 3}~$
DC's of $~\vec{CQ}~$ are $~\dfrac{1}{\sqrt 3}~,-\dfrac{1}{\sqrt 3},~-\dfrac{1}{\sqrt 3}~$
Let $~l,~m,~n~$ be DC's of line $($and hence $~l^2+m^2+n^2=1~)$ and line makes angle $~α~$ with $~\vec{OP}~$. So
$$\cos α=l\left(\dfrac{1}{\sqrt 3}\right)+m\left(\dfrac{1}{\sqrt 3}\right)+n\left(\dfrac{1}{\sqrt 3}\right)=\dfrac{l+m+n}{\sqrt 3}$$
Similarly $$\cos \beta=l\left(-\dfrac{1}{\sqrt 3}\right)+m\left(\dfrac{1}{\sqrt 3}\right)+n\left(\dfrac{1}{\sqrt 3}\right)=\dfrac{-l+m+n}{\sqrt 3}~,$$
$$\cos \gamma=l\left(\dfrac{1}{\sqrt 3}\right)+m\left(-\dfrac{1}{\sqrt 3}\right)+n\left(\dfrac{1}{\sqrt 3}\right)=\dfrac{l-m+n}{\sqrt 3}~,$$
$$\cos \delta=l\left(\dfrac{1}{\sqrt 3}\right)+m\left(\dfrac{1}{\sqrt 3}\right)+n\left(-\dfrac{1}{\sqrt 3}\right)=\dfrac{l+m-n}{\sqrt 3}$$
Now
\begin{equation}
\cos2\alpha+\cos2\beta+\cos2\gamma+\cos2\delta\\
=2\left(\cos^2\alpha+\cos^2\beta+\cos^2\gamma+\cos^2\delta\right)-4\\
=\dfrac 23\left[(l+m+n)^2+(-l+m+n)^2+(l-m+n)^2+(l+m-n)^2\right]-4\\
=\dfrac 23\left[4(l^2+m^2+n^2)\right]-4\\
=4\left[\dfrac 23-1\right]\\
=-\dfrac 43
\end{equation}
In general, the angle between a line with direction $\mathbf v$ and a plane with normal $\mathbf n$ is $$\theta=\arcsin\left|\frac{\mathbf v \cdot\mathbf n}{\left|\mathbf v\right| \,\left|\mathbf n\right|} \right|.$$
Since the $xy-,$ $yz-, zx-$ planes have respective normals $\begin{pmatrix}0\\0\\1\end{pmatrix},\begin{pmatrix}1\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\end{pmatrix},$ as pointed out by Ted, your own answers are indeed correct.
P.S. The above formula has a similar derivation as yesterday's formula for the angle between two planes.
Best Answer
We have that
$$\vec r=(r_x,r_y,r_z)=k(2,3,-6)$$
and
$$|\vec r|=k\sqrt{4+9+36}=7k=14 \implies k=2$$
therefore
$$\vec r=(4,6,-12)\implies \frac{\vec r}{|\vec r|}=\left(\frac27,\frac37,-\frac67\right)$$