Prove that the paraboloids have a common tangent planes

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The discussion focuses on proving that three specific paraboloids have a common tangent plane under the condition that the determinant of the matrix formed by their coefficients equals zero. The paraboloids are defined by the equations $$\frac{x^2}{a_1^2}+\frac{y^2}{b_1^2}=\frac{2z}{c_1}$$, $$\frac{x^2}{a_2^2}+\frac{y^2}{b_2^2}=\frac{2z}{c_2}$$, and $$\frac{x^2}{a_3^2}+\frac{y^2}{b_3^2}=\frac{2z}{c_3}$$. The necessary condition for the common tangent plane is established through the determinant $$\begin{bmatrix}a_1^2 & a_2^2 & a_3^2\\ b_1^2 & b_2^2 & b_3^2\\ c_1 & c_2 & c_3\end{bmatrix}=0$$, where all parameters are non-zero real numbers.

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Prove that the paraboloids:
$$\frac{x^2}{a_1^2}+\frac{y^2}{b_1^2}=\frac{2z}{c_1}$$;

$$\frac{x^2}{a_2^2}+\frac{y^2}{b_2^2}=\frac{2z}{c_2}$$;

$$\frac{x^2}{a_3^2}+\frac{y^2}{b_3^2}=\frac{2z}{c_3}$$

Have a common tangent plane if:
$$\begin{bmatrix}a_1^2 & a_2^2 & a_3^2\\ b_1^2 & b_2^2 & b_3^2\\ c_1 & c_2 & c_3\end{bmatrix}=0$$
Here
$$a_i, b_i, c_i \in \Bbb{R} \left\{0\right\}$$
 
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Hello debrajr and welcome to MHB! :D

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

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