Phương pháp giải:

$V = \frac{1}{6}\left| {\overrightarrow {AD} .\left[ {\overrightarrow {AB} ,\overrightarrow {AC} } \right]} \right|$

Lời giải chi tiết:

$\overrightarrow {AB}  = \left( {1; - 1;2} \right),\overrightarrow {AC}  = \left( {0; - 2;4} \right)$

$\left[ {\overrightarrow {AB} ,\overrightarrow {AC} } \right] = \left( {0; - 4; - 2} \right)$

$D \in Oy \Rightarrow D\left( {0;y;0} \right)$$\Rightarrow \overrightarrow {AD}  = \left( { - 2;y - 1;1} \right)$

$\begin{array}{l}V = \frac{1}{6}\left| {\overrightarrow {AD} .\left[ {\overrightarrow {AB} ,\overrightarrow {AC} } \right]} \right| = 5\\ \Leftrightarrow \left| {\overrightarrow {AD} .\left[ {\overrightarrow {AB} ,\overrightarrow {AC} } \right]} \right| = 30\end{array}$

Nếu $\overrightarrow {AD} .\left[ {\overrightarrow {AB} ,\overrightarrow {AC} } \right] = 30$ thì

$\begin{array}{l}\left( { - 2} \right).0 + \left( {y - 1} \right).\left( { - 4} \right) + 1.\left( { - 2} \right) = 30\\ \Leftrightarrow y =  - 7 =  > D\left( {0; - 7;0} \right)\end{array}$

Nếu $\overrightarrow {AD} .\left[ {\overrightarrow {AB} ,\overrightarrow {AC} } \right] =  - 30$ thì

$\begin{array}{l}\left( { - 2} \right).0 + \left( {y - 1} \right).\left( { - 4} \right) + 1.\left( { - 2} \right) =  - 30\\ \Leftrightarrow y = 8 \Rightarrow D\left( {0;8;0} \right)\end{array}$