Explanation

(d)
Given vertices are
$A\left( 3\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}} \right),\ B\left( \ \widehat{\dot{\mathbf{i}}} - 2\widehat{\dot{\mathbf{j}}} + 7\widehat{\mathbf{k}} \right)andC\left( - 2\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} + 5\widehat{\mathbf{k}} \right).$
Now, $\overrightarrow{\mathbf{AB}} = \left( \widehat{\dot{\mathbf{i}}} - 2\widehat{\dot{\mathbf{j}}} + 7\widehat{\mathbf{k}} \right) - \left( 3\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}} \right)$
$= - 2\widehat{\dot{\mathbf{i}}} - 3\widehat{\dot{\mathbf{j}}} + 5\widehat{\mathbf{k}}$
$\therefore\ \left| \overrightarrow{\mathbf{AB}} \right| = \sqrt{4 + 9 + 25} = \sqrt{38}$
Similarly, $\left| \overrightarrow{\mathbf{BC}} \right| = \left| \overrightarrow{\mathbf{CA}} \right| = \sqrt{38}$
$\therefore\ \left| \overrightarrow{\mathbf{AB}} \right| = \left| \overrightarrow{\mathbf{BC}} \right| = \left| \overrightarrow{\mathbf{CA}} \right| = \sqrt{38}$
$\therefore\ \$Hence, triangle is an equilateral triangle.