Consider the plane
 $E:\overrightarrow{r}=\left[\begin{array}{c}-1\\ 1\\ 1\end{array}\right]+\lambda \left[\begin{array}{c}1\\ 2\\ 0\end{array}\right]+\mu \left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]$
Let F be the plane containing the point A(-4, 2, 2) and parallel to E. Suppose the point B is on the plane E such that B has a minimum distance from the point A if C(-3, 0, 4) lies in the plane F. If A is area of the angle ABC then find [A] (where [ ] denotes GIF).
 

 $$\begin{gathered} E:\overline{r}=(–1+j+k)+\lambda (i+2j+0k)+\mu (i+0j+k) \\ \overrightarrow{n}=\left|\begin{array}{ccc}i& j& k\\ 1& 2& 0\\ 1& 0& 1\end{array}\right| \\ =i\left(2\right)–j\left(1\right)+\widehat{k}(–2) \\ \overrightarrow{n}=2i–j–2k \end{gathered}$$
2(x + 4) – 1(y – 2) – 2 (z – 2) = 0
2x – y – 2z + 8 + 2 + 4 = 0
2x – y – 2z + 14 = 0
$$\begin{gathered} F:\overrightarrow{r}.(2i–j–2k)=–14 \\ E:\overrightarrow{r}.(2i–j–2k)=–5 \\ \overrightarrow{E}\Vert \overrightarrow{F} \\ \overrightarrow{r}=(–4i+2j+2k)+\lambda (2i–j–2k) \\ B(–4+2\lambda 2–\lambda 2–2\lambda ) \end{gathered}$$ 
  B lie on plane  $\overrightarrow{r}.(2i–j–2\widehat{k})=–5$
 
 $$\begin{gathered} 2(–4+2\lambda )–(2–\lambda )–2(2–2\lambda )=–5 \\ 8+4\lambda –2+\lambda –4+4\lambda =–5 \\ 9\lambda =9 \\ \lambda =1 \end{gathered}$$
 $$\begin{gathered} B(–2,1,0) \\ A(–4,2,2)C(–3,0,4) \\ \overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}i& j& k\\ 2& –1& –2\\ 1& –2& 2\end{array}\right| \\ =i(–6)–j\left(6\right)–3k=–6i–6j–3k \\ \frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|=\frac{1}{2}\sqrt{36+36+9} \\ =\frac{\sqrt{81}}{2}=\frac{9}{2} \end{gathered}$$