Answer
An equation of the plane is $10x+15y+6z=30$.
Work Step by Step
Let $P$, $Q$ and $R$ denote the intersections points of the plane with $x$-axis, $y$-axis and $z$-axis, respectively. From Figure 8 we obtain
$P = \left( {3,0,0} \right)$, ${\ \ }$ $Q = \left( {0,2,0} \right)$, ${\ \ }$ $R = \left( {0,0,5} \right)$
Write the vector:
${\bf{u}} = \overrightarrow {PQ} = Q - P = \left( {0,2,0} \right) - \left( {3,0,0} \right) = \left( { - 3,2,0} \right)$.
Write the vector:
${\bf{v}} = \overrightarrow {PR} = R - P = \left( {0,0,5} \right) - \left( {3,0,0} \right) = \left( { - 3,0,5} \right)$.
Since ${\bf{u}}$ and ${\bf{v}}$ lie in the plane, the cross product of them form a vector normal to the plane. That is,
${\bf{n}} = {\bf{u}} \times {\bf{v}} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{ - 3}&2&0\\
{ - 3}&0&5
\end{array}} \right|$
$\left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{ - 3}&2&0\\
{ - 3}&0&5
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
2&0\\
0&5
\end{array}} \right|{\bf{i}} - \left| {\begin{array}{*{20}{c}}
{ - 3}&0\\
{ - 3}&5
\end{array}} \right|{\bf{j}} + \left| {\begin{array}{*{20}{c}}
{ - 3}&2\\
{ - 3}&0
\end{array}} \right|{\bf{k}}$
So, ${\bf{n}} = 10{\bf{i}} + 15{\bf{j}} + 6{\bf{k}}$
By Eq. (5) of Theorem 1, the equation of the plane is given by
$10x + 15y + 6z = d$
We choose $P = \left( {3,0,0} \right)$ and substituting it in the equation above and obtain $d=30$. Thus, the equation of the plane is $10x+15y+6z=30$.
