The area of the quadrilateral $\mathrm{ABCD}$ with vertices $\mathrm{A}(2,1,1), \mathrm{B}(1,2,5), \mathrm{C}(-2,-3,5)$ and $\mathrm{D}(1,-6,-7)$ is equal to :

$$ \begin{aligned} & \text { Here } \overrightarrow{\mathrm{AC}}=(-2-2) \hat{i}+(-3-1) \hat{j}+(5-1) \hat{k} \\\\ & =-4 \hat{i}-4 \hat{j}+4 \hat{k} \\\\ & \overrightarrow{\mathrm{BD}}=(1-1) \hat{i}+(-6-2) \hat{j}+(-7-5) \hat{k} \\\\ & =-8 \hat{j}-12 \hat{k} \end{aligned} $$ <br/><br/>So, area of quadrilateral $=\frac{1}{2}| \overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}} \mid$ <br/><br/>$$ \begin{aligned} & =\frac{1}{2}\left\|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -4 & -4 & 4 \\ 0 & -8 & -12 \end{array}\right\| \\\\ & =\frac{1}{2}|(48+32) \hat{i}-(48-0) \hat{j}+(32-0) \hat{k}| \\\\ & =\frac{1}{2}|80 \hat{i}-48 \hat{j}+32 \hat{k}| \\\\ & =\frac{1}{2} 16|15 \hat{i}-3 \hat{j}+2 \hat{k}| \\\\ & =8 \sqrt{25+9+4}=8 \sqrt{38} \text { sq units. } \end{aligned} $$