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The normal  
\[\vec{n}\]
  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
\[\begin{equation} \begin{aligned} d &=\sqrt{(\frac{a}{a^2+b^2+c^2} -0)^2+(\frac{b}{a^2+b^2+c^2} -0)^2+(\frac{c}{a^2+b^2+c^2} -0)^2} \\ &= \sqrt{\frac{d^2(a^2+b^2+c^2)}{(a^2+b^2+c^2)^2}} \\ &= \frac{d}{\sqrt{a^2+b^2+c^2}}\end{aligned} \end{equation} \]