Let $\mathrm{AB}$ makes angles, $\alpha, \beta$ and $\gamma$ with $O X, O Y$ and $O Z$, respectively. Then, $\mathrm{AB}=d$ $(\cos \alpha \hat{\hat{i}}+\cos \beta \hat{\mathbf{j}}+\cos \gamma \hat{\mathbf{k}})=d \cos \alpha \hat{\hat{i}}+\mathrm{d} \cos \beta \hat{\mathbf{j}}+\mathrm{d} \cos \gamma \hat{\mathbf{k}}$
Let $\theta$ be the angle between $\overrightarrow{\mathbf{A B}}$ and $X Y$ - plane, Then, $d_1=d \cos \theta$




$$
\text { Here, } \begin{aligned}
\sin \theta & =\frac{\mathbf{A B} \cdot \hat{\mathbf{k}}}{|\mathbf{A B}||\hat{\mathbf{k}}|} \\
& =\frac{d \cos \gamma}{d \mathbf{l}}=\cos \gamma=\sin \left(\frac{\pi}{2}-\gamma\right) \\
\Rightarrow \quad \theta & =\frac{\pi}{2}-\gamma \\
\therefore \quad d_1 & =d \cos \left(\frac{\pi}{2}-\gamma\right)=d \sin \gamma
\end{aligned}
$$
Similarly, $d_2=$ projection of $\mathbf{A B}$ on $Y Z$ - plane
$$
=d \sin \alpha
$$
and $d_3=$ project of $\mathbf{A B}$ on $2 X X$-plane
$$
=d \sin \beta
$$
We know that, $\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1$
$$
\begin{array}{rr}
\therefore 1-\sin ^2 \alpha+1-\sin ^2 \beta+1-\sin ^2 \gamma=1 \\
\Rightarrow & \sin ^2 \alpha+\sin ^2 \beta+\sin ^2 \gamma=2 \\
\Rightarrow & \left(\frac{d_2}{d}\right)^2+\left(\frac{d_3}{d}\right)^2+\left(\frac{d_1}{d}\right)^2=2
\end{array}
$$
[using Eqs. (i), (ii) and (iii)]
$$
\begin{array}{ll}
\Rightarrow & \frac{d_1^2}{d^2}+\frac{d_2^2}{d^2}+\frac{d_3^2}{d^2}=2 \\
\Rightarrow & d_1^2+d_2^2+d_3^2=2 d^2
\end{array}
$$