Explanation

(c)
Let $\overrightarrow{\mathbf{A}} = 7\widehat{\dot{\mathbf{j}}} + 10\widehat{\mathbf{k}},\ \ \overrightarrow{\mathbf{B}} = - \widehat{\dot{\mathbf{i}}} + 6\widehat{\dot{\mathbf{j}}} + 6\widehat{\mathbf{k}}\$and$\ \overrightarrow{\mathbf{C}} = - 4\widehat{\dot{\mathbf{i}}} + 9\widehat{\dot{\mathbf{j}}} + 6\widehat{\mathbf{k}}$
$Now,\ \overrightarrow{\mathbf{AB}} = - \widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} - 4\widehat{\mathbf{k}},\ \ \overrightarrow{\mathbf{BC}} = - 3\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}}$
$and\ \overrightarrow{\mathbf{CA}} = 4\widehat{\dot{\mathbf{i}}} - 2\widehat{\dot{\mathbf{j}}} + 4\widehat{\mathbf{k}}$
$Here,\ \left| \overrightarrow{\mathbf{AB}} \right| = \left| \overrightarrow{\mathbf{BC}} \right| = 3\sqrt{2}$ and |$\overrightarrow{\mathbf{CA}}| = 6$
Now, $|\overrightarrow{\mathbf{AB}}|^{2} + |\overrightarrow{\mathbf{BC}}|^{2} = |\overrightarrow{\mathbf{AC}}|^{2}$
Hence, the triangle is right angled isosceles triangle.