\[\vec{n}\]
to the plane \[ax+by+cz=d\]
is \[\begin{pmatrix}a\\b\\c\end{pmatrix}\]
. The line
\[r(t)=\begin{pmatrix}a\\b\\c\end{pmatrix} t\]
is perpendicular to the plane and passes through the origin passes through the origin. For this line, the \[x, \; y, \; z\]
coordinates are \[at, \; bt, \; ct\]
respectively. Substitute these into the equation of the plane.\[a(at)+b(bt)+c)ct)=d \rightarrow t=\frac{d}{a^2+b^2+c^2}\]
.Substitute this value of
\[t\]
into the equation of the line to get the coordinate of the intersection of the line with the plane.\[\begin{pmatrix}a\\b\\c\end{pmatrix}^T \frac{d}{a^2+b^2+c^2}= (\frac{a}{a^2+b^2+c^2}, \frac{b}{a^2+b^2+c^2}, \frac{c}{a^2+b^2+c^2} ) \]
.The distance between this point and the origin
\[(0,0,0)\]
is\[\begin{equation} \begin{aligned} d &=\sqrt{(\frac{a}{a^2+b^2+c^2} -0)^2+(\frac{b}{a^2+b^2+c^2} -0)^2+(\frac{c}{a^2+b^2+c^2} -0)^2} \\ &= \sqrt{\frac{d^2(a^2+b^2+c^2)}{(a^2+b^2+c^2)^2}} \\ &= \frac{d}{\sqrt{a^2+b^2+c^2}}\end{aligned} \end{equation} \]