Prove that the paraboloids $\frac{x^2}{a_{1}^2} + \frac{y^2}{b_{1}^2} = \frac{2z}{c_{1}}$, $\frac{x^2}{a_{2}^2} + \frac{y^2}{b_{2}^2} = \frac{2z}{c_{2}}$ and $\frac{x^2}{a_{3}^2} + \frac{y^2}{b_{3}^2} = \frac{2z}{c_{3}}$ have a common tangent plane if $$det\begin{pmatrix}a_{1}^2&a_{2}^2&a_{3}^2\\b_{1}^2&b_{2}^2&b_{3}^2\\c_{1}&c_{2}&c_{3}\end{pmatrix} = 0$$
If $(x_{0}, y_{0}, z_{0})$ is the point of tangency, then I know that the tangent planes to each of the paraboloids are of the form $\frac{xx_{0}}{a_{1}^2} + \frac{yy_{0}}{b_{1}^2} = \frac{z + z_{0}}{c_{1}}$, $\frac{xx_{0}}{a_{2}^2} + \frac{yy_{0}}{b_{2}^2} = \frac{z + z_{0}}{c_{2}}$ and $\frac{xx_{0}}{a_{3}^2} + \frac{yy_{0}}{b_{3}^2} = \frac{z + z_{0}}{c_{3}}$
If the three planes are coincident, the equations will be scalar multiples of each other. However, this doesn't seem to help in getting the given condition.
Any help would be greatly appreciated.
Since the equations of the tangent planes are scalar multiples of each other, $a_{1}^2$, $b_{1}^2$, $c_{1}$ are proportional to $a_{2}^2$, $b_{2}^2$, $c_{2}$ and $a_{3}^2$, $b_{3}^2$, $c_{3}$.
The columns of $\begin{pmatrix}a_{1}^2&a_{2}^2&a_{3}^2\\b_{1}^2&b_{2}^2&b_{3}^2\\c_{1}&c_{2}&c_{3}\end{pmatrix}$ are proportional to each other , therefore the determinant is zero.