Explanation

(d)
The position vector of the vertices $A,B,C$ of $\mathrm{\Delta}ABC$ are $7\widehat{\mathbf{j}} + 10\widehat{\mathbf{k}},\ - \widehat{\mathbf{i}} + 6\widehat{\mathbf{j}} + 6\widehat{\mathbf{k}}$ and $- 4\widehat{\mathbf{i}} + 9\widehat{\mathbf{j}} + 6\widehat{\mathbf{k}}$ respectively.
$\therefore\ \overrightarrow{\mathbf{AB}} = - \widehat{\mathbf{i}} - \widehat{\mathbf{j}} - 4\widehat{\mathbf{k}}$, $\overrightarrow{\mathbf{BC}} = - 3\widehat{\mathbf{i}} + 3\widehat{\mathbf{j}}$
And $\overrightarrow{\mathbf{CA}} = 4\widehat{\mathbf{i}} - 2\widehat{\mathbf{j}} - 4\widehat{\mathbf{k}}$
$\Rightarrow |\ \overrightarrow{\mathbf{AB}}| = \sqrt{( - 1)^{2} + ( - 1)^{2} + ( - 4)^{2}} = \sqrt{18} = 3\sqrt{2}$
$|\overrightarrow{\mathbf{BC}}| = \sqrt{( - 3)^{2} + 3^{2}} = \sqrt{18} = 3\sqrt{2}$
and $\left| \ \overrightarrow{\mathbf{CA}} \right| = \sqrt{4^{2} + ( - 2)^{2} + ( - 4)^{2}} = \sqrt{36} = 6$
It is clear from these values that
$\left| \ \overrightarrow{\mathbf{AB}}\left. \ \right|^{2} + \right|\overrightarrow{\mathbf{BC}}\left. \ \right|^{2}\ = \left| \ \overrightarrow{\mathbf{CA}} \right|^{2}$
Hence, $\mathrm{\Delta}\ ABC$ is right angled and isosceles also.