ABC is a triangle, the position vectors of whose vertices are $7\widehat{\mathrm{j}}+10\widehat{\mathrm{k}},-\widehat{\mathrm{i}}+6\widehat{\mathrm{j}}+6\widehat{\mathrm{k}}$ and $-4\widehat{i}+9\widehat{j}+6\widehat{k}$ then $∆$ABC is

we have $\overrightarrow{\mathrm{AB}}=-\widehat{\mathrm{i}}-\widehat{\mathrm{j}}-4\widehat{\mathrm{k}},\overrightarrow{\mathrm{BC}}=-3\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}$ and $\overrightarrow{\mathrm{CA}}=4\widehat{\mathrm{i}}-2\widehat{\mathrm{j}}+4\widehat{\mathrm{k}}$

Therefore, $\left|\overrightarrow{AB}\right|=\left|\overrightarrow{BC}\right|=3\sqrt{2}$

clearly ,$|\overrightarrow{\mathrm{AB}}{|}^2+|\overrightarrow{\mathrm{BC}}{|}^2=|\overrightarrow{\mathrm{AC}}{|}^2$

Hence, the triangle is right-angled isosceles triangle.