9. Ypannenne npassoR, npoxozaueh 4epe3 Touky M_(0)(1;0;-2) napanzenbHo BeKTOPY s=(1;-2;3), HMCET BHA: 1) (x-1)/(1)=(y)/(-2)=(z+2)/(3) 2) (x-1)/(1)=(y+2)/(0)=(z-2)/(-2) 3) (x-1)/(1)=(y)/(-2)=(z-2)/(3) 4) (x-1)/(1)=(y)/(2)=(z+2)/(3)

To find the equation of the plane passing through the point $M_{0}(1;0;-2)$ and perpendicular to the vector $s=(1;-2;3)$, we can use the formula for the equation of a plane in three-dimensional space.<br /><br />The equation of a plane passing through a point $(x_{0}, y_{0}, z_{0})$ and perpendicular to a vector $s=(a, b, c)$ is given by:<br /><br />$\frac{x-x_{0}}{a}=\frac{y-y_{0}}{b}=\frac{z-z_{0}}{c}$<br /><br />In this case, the point is $M_{0}(1;0;-2)$ and the vector is $s=(1;-2;3)$. Substituting these values into the formula, we get:<br /><br />$\frac{x-1}{1}=\frac{y-0}{-2}=\frac{z+2}{3}$<br /><br />Simplifying this equation, we get:<br /><br />$\frac{x-1}{1}=\frac{y}{-2}=\frac{z+2}{3}$<br /><br />Therefore, the correct answer is option 1) $\frac{x-1}{1}=\frac{y}{-2}=\frac{z+2}{3}$.