A Tetrahedron $ABCD$ having its face $ABD,BCD$
$CDA,ABC$ parallel to the plane $x+y=2,y+z=2,$
$z+x=2,x+y+z=4$ respectively.
If $AB=AC=\sqrt{2},AD=\sqrt{3}$ unit.
Then volume of Tetrahedron in cubic unit is
what i try
solving
$\left.\begin{matrix} x+y=2 & \\ x+y+z=4 & \end{matrix}\right\}\Rightarrow z=2$
$\left.\begin{matrix} y+z=2 & \\ x+y+z=4 & \end{matrix}\right\}\Rightarrow x=2$
$\left.\begin{matrix} x+z=2 & \\ x+y+z=4 & \end{matrix}\right\}\Rightarrow y=2$
Let position vector of point $D(\vec{0}),A(\vec{a}),B(\vec{b}),C(\vec{c})$
Then volume of Tetrahedron whose edges are $\vec{a},\vec{b},\vec{c}$ is
$$\frac{1}{6}\bigg|[\vec{a}\vec{b}\vec{c}]\bigg|\;\text{cubic unit}$$
How do i solve it Help me please
Let us first recall that a plane with equation $ux+vy+wz=h$ has the vector with coordinates $(u,v,w)$ as its normal vector.
Consider figure below (coordinates of the vectors normal to the different faces are obtained from your equations):
Fig. 1 : The names of normal vectors have been attributed accordingly to the name of the opposite vertex. Please note that the orientation of these normals (inward/outward) is not necessarily the good one.
As a consequence :
$$\vec{AB}= \alpha \ c \times d= \alpha \ \begin{pmatrix}1\\-1\\0\end{pmatrix}, \ \ \vec{AC}= \beta \ d \times b=\beta \ \begin{pmatrix}1\\0\\-1\end{pmatrix}, \ \ \vec{AD}= \gamma \ b \times c=\gamma \ \begin{pmatrix}1\\-1\\1\end{pmatrix}, \ \ $$
(we can take normalization constants $\alpha=\beta=\gamma=1$ : we are fortunate they are in agreement with the length values $\sqrt{2}$ and $\sqrt{3}$ !).
Therefore, the volume of the tetrahedron is given by :
$$V= \dfrac16 \left|\begin{array}{rrr} 1&1&1\\-1&0&-1\\0&-1&1\end{array}\right|=\dfrac16$$