Find the normal vector to the plane defined by the equation x − 2y + z − 5 = 0.

The general equation of a plane is given by $Ax + By + Cz + D = 0$.

The coefficients of $x$, $y$, and $z$ in the plane equation directly form the components of the normal vector to the plane. The normal vector is $\vec{n} = \begin{pmatrix} A \\ B \\ C \end{pmatrix}$.

The given equation is $x - 2y + z - 5 = 0$.
Comparing this to $Ax + By + Cz + D = 0$:
$A = 1$ (coefficient of $x$)
$B = -2$ (coefficient of $y$)
$C = 1$ (coefficient of $z$)

Using the coefficients, the normal vector is:

The vector normal to this plane is: