Find x such that the four points A(4, 4, 4), B(5, x, 8), C(5, 4, 1) and D (7, 7, 2) are coplanar. 

OR

Using vectors, find the area of the ΔABC, whose vertices are A(1, 2, 5), 5(2, -1, 4) and C(4, 5, -1).

Given points are A(4,4,4), B(5, x, 8), C(5,4,1) and D(7,7,2), then position vectors of A, B, C and D respectively, are
$\begin{array}{l}\overrightarrow{\mathrm{OA}}=4\widehat{\mathrm{i}}+4\widehat{\mathrm{j}}+4\widehat{\mathrm{k}},\overrightarrow{\mathrm{OB}}=5\widehat{\mathrm{i}}+\mathrm{x}\widehat{\mathrm{j}}+8\widehat{\mathrm{k}}\\ \overrightarrow{\mathrm{OC}}=5\widehat{\mathrm{i}}+4\widehat{\mathrm{j}}+\widehat{\mathrm{k}}\text{ and }\overrightarrow{\mathrm{OD}}=7\widehat{\mathrm{i}}+7\widehat{\mathrm{j}}+2\widehat{\mathrm{k}}\\ \therefore \overrightarrow{\mathrm{AB}}=(5\widehat{\mathrm{i}}+\mathrm{x}\widehat{\mathrm{j}}+8\widehat{\mathrm{k}})-(4\widehat{\mathrm{i}}+4\widehat{\mathrm{j}}+4\widehat{\mathrm{k}})\\ =\widehat{\mathrm{i}}+(\mathrm{x}-4)\widehat{\mathrm{j}}+4\widehat{\mathrm{k}}\\ \overrightarrow{\mathrm{AC}}=(5\widehat{\mathrm{i}}+4\widehat{\mathrm{j}}+\widehat{\mathrm{k}})-(4\widehat{\mathrm{i}}+4\widehat{\mathrm{j}}+4\widehat{\mathrm{k}})\\ =\widehat{\mathrm{i}}-3\widehat{\mathrm{k}}\\ \text{ and }\overrightarrow{\mathrm{AD}}=(7\widehat{\mathrm{i}}+7\widehat{\mathrm{j}}+2\widehat{\mathrm{k}})-(4\widehat{\mathrm{i}}+4\widehat{\mathrm{j}}+4\widehat{\mathrm{k}})\\ =3\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}-2\widehat{\mathrm{k}}\end{array}$
Given points are coplanar, if vectors $\overrightarrow{\mathrm{AB}},\overrightarrow{\mathrm{AC}},\overrightarrow{\mathrm{AD}}$ are coplanar.
$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}[\overrightarrow{\mathrm{AB}} \overrightarrow{\mathrm{AC} }\overrightarrow{\mathrm{AD}}]=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left|\begin{array}{ccc}1& \mathrm{x}-4& 4\\ 1& 0& -3\\ 3& 3& -2\end{array}\right|=0\end{array}$
$\begin{array}{l}\Rightarrow 1(0+9)-(\mathrm{x}-4)(-2+9)+4(3-0)=0\\ \Rightarrow 9-(\mathrm{x}-4)\left(7\right)+12=0\\ \Rightarrow 9-7\mathrm{x}+28+12=0\\ \Rightarrow 49-7\mathrm{x}=0\\ \Rightarrow 7\mathrm{x}=49\Rightarrow \mathrm{x}=7\end{array}$

OR

Let the position vectors of the verices A, B and C of $△$A B C be
$\overrightarrow{\mathrm{OA}}=\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+3\widehat{\mathrm{k}},\overrightarrow{\mathrm{OB}}=2\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+4\widehat{\mathrm{k}}$
$\text{ and }\overrightarrow{\mathrm{OC}}=4\widehat{\mathrm{i}}+5\widehat{\mathrm{j}}-\widehat{\mathrm{k}}\text{, respectively. }$

Then, $\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}$
$\begin{array}{l}=(2\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+4\widehat{\mathrm{k}})-(\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+3\widehat{\mathrm{k}})\\ =\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}+\widehat{\mathrm{k}}\\ \text{ and }\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}\\ =(4\widehat{\mathrm{i}}+5\widehat{\mathrm{j}}-\widehat{\mathrm{k}})-(\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+3\widehat{\mathrm{k}})\\ =(3\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}-4\widehat{\mathrm{k}})\\ \text{ Now, }\overrightarrow{\mathrm{AB}}\times \overrightarrow{\mathrm{AC}}=\left|\begin{array}{ccc}1& -3& 1\\ 3& 3& -4\end{array}\right|\\ =\widehat{\mathrm{i}}(12-3)-\widehat{\mathrm{j}}(-4-3)+\widehat{\mathrm{k}}(3+9)\\ =9\widehat{\mathrm{i}}+7\widehat{\mathrm{j}}+12\widehat{\mathrm{k}} \\ \end{array}$
$\begin{array}{l}|\overrightarrow{\mathrm{AB}}\times \overrightarrow{\mathrm{AC}}|=\sqrt{(9{)}^2+(7{)}^2+(12{)}^2}=\sqrt{81+49+144}=\sqrt{274}\\ \therefore \text{ Area of }\mathrm{△}\mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}}\times \overrightarrow{\mathrm{AC}}|\\ =\frac{1}{2}\sqrt{274} \mathrm{squnits}\\ \end{array}\mathrm{ }$