A(2,0,0) , B(0,3,0) , C(0,0,6) , D(2,3,8) are the vertices…

$A(2,0,0)$, $B(0,3,0)$, $C(0,0,6)$, $D(2,3,8)$ are the vertices of a pyramid. $1^\circ$ Compute its volume. $2^\circ$ Compute the height onto the base $ABC$.

Step 1: Edge vectors from $A$

\[ \overrightarrow{AB}=\{-2,3,0\},\qquad \overrightarrow{AC}=\{-2,0,6\},\qquad \overrightarrow{AD}=\{0,3,8\}. \]

Step 2: $1^\circ$ Volume

\[ [\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD}]=\begin{vmatrix}-2&3&0\\-2&0&6\\0&3&8\end{vmatrix}=-2(0-18)-3(-16-0)+0=36+48=84. \]

\[V=\tfrac16\cdot84=14\]

Step 3: $2^\circ$ Height

\[ \overrightarrow{AB}\times\overrightarrow{AC}=\{18,12,6\},\qquad |\overrightarrow{AB}\times\overrightarrow{AC}|=\sqrt{504}=6\sqrt{14},\qquad S_{ABC}=3\sqrt{14}, \]

\[ h=\frac{3V}{S_{ABC}}=\frac{42}{3\sqrt{14}}=\frac{14}{\sqrt{14}}. \]

\[h=\sqrt{14}\approx3.74\]

Step 4: Independent check (added)

The intercepts $2,3,6$ give the plane $\tfrac x2+\tfrac y3+\tfrac z6=1$, i.e. $3x+2y+z=6$. Its normal is $\{3,2,1\}$, parallel to $\overrightarrow{AB}\times\overrightarrow{AC}$. The distance of $D$ from it is

\[ h=\frac{|3\cdot2+2\cdot3+8-6|}{\sqrt{9+4+1}}=\frac{14}{\sqrt{14}}=\sqrt{14}\ \checkmark,\qquad H=D-\{3,2,1\}=(-1,1,7). \]