Question

Does there exist a plane targent to x²-2y²-2z²=8 and which passes through
2x+3y+2z = 8, x−y+2z = 5? Justify your answer.

Answer 1

The equation of the straight line is

$2x+3y+2z=8\\ x-y+2z=5$

the direction vector

$\vec{a}=[\vec{n_1},\vec{n_2}]\\ \vec{n_1}=(2;3;2)\\ \vec{n_2}=(1;-1;2)\\ \vec{a}=\left( \begin{vmatrix} 3 & 2 \\ -1 & 2 \end{vmatrix};-\begin{vmatrix} 2 & 2 \\ 1 & 2 \end{vmatrix};\begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix}\right)=\\ =(8;-2-5)$

 point

let $y=0$ , then

$2x+2z=8\\ x+2z=5\\ x=3\\ z=1$

$A(3,0,1)$

The equation of a straight line is

$\frac{x-3}{8}=\frac{y-0}{-2}=\frac{z-1}{-5}$

Let's take an arbitrary point $M(x_0;y_0;z_0)$ on the surface

$x^2-2y^2-2z^2=8\\ x_0^2-2y_0^2-2z_0^2=8\\$

 The equation of a plane is

$\begin{vmatrix} x-x_0 & y-y_0&z-z_0 \\ 3-x_0 & 0-y_0& 1-z_0\\ 8&-2&-5 \end{vmatrix}=0$  

$(x-x_0)(5y_0+2-2z_0)+(y-y_0)(15-5x_0+8-8z_0)+\\ +(z-z_0)(-6+2x_0+8y_0)=0\\ (x-x_0)(5y_0-2z_0+2)+(y-y_0)(-5x_0-8z_0+23)+\\ +(z-z_0)(2x_0+8y_0-6)=0\\$

hence the plane exists.