The position vectors of the vertices A,B,C of a tetrahedron ABCD are $\widehat{i}+\widehat{j}+\widehat{k},\widehat{i}$ and $3\widehat{i}$ respectively. The altitude from the vertex D to the opposite face ABC meet the median line through A of the at E. If the length of side AD is 4 and volume of tetrahedron is $\frac{2\sqrt{2}}{3}$ then position vector of E is

$\text{ Line }\frac{x-0}{2}=\frac{y-2}{2}=\frac{z-1}{0}=r(2r,2r+2,1)$

Point lie on $x=0\Rightarrow r=0$

Point A(0,2,1)

$\text{ Let }B(-2\alpha ,\alpha ,0)$

Image of $B(-2\alpha ,\alpha ,\beta )$ w.r.t x = 0 plane is
Direction ratio of AB
is $2\alpha ,\alpha -2,\beta -1$ is
proportional to 2,2,0

$\frac{2\alpha }{2}=\frac{\alpha -2}{2}=\frac{\beta -1}{0}\Rightarrow \beta =1,\alpha =-2$

Point B(4, –2, 1) and B(
4,
2,1)
image of A(0,2,1) with respect to x + 2y = 0

$\frac{x-0}{1}=\frac{y-2}{2}=\frac{z-0}{0}=\frac{-2(0+4)}{(1+4)}=-\frac{8}{5}$

$\Rightarrow A^{\mathrm{\prime }}\left(\frac{-8}{5},\frac{-8}{5},1\right)$

Equation of $A^{\mathrm{\prime }}B\frac{x-4}{-\frac{28}{5}}=\frac{y+2}{\frac{4}{5}}=\frac{z-1}{0}=r$

$\text{ Any point }\left(\frac{-28}{5}r+4,\frac{4r}{5}-2,1\right)$ lie on line

$\frac{x-0}{2}=\frac{y-2}{2}=\frac{z-1}{0}$

$\frac{\frac{-28r}{5}+4}{2}=\frac{\frac{4r}{5}-2-2}{2}\Rightarrow 8=\frac{4r}{5}+\frac{28r}{5}\Rightarrow r=\frac{5}{4}$

$\therefore \text{ Points }J\text{ is }(-3,-1,1)$