Example 1.23. Find the equation of the plane passing through the points $M(2, -1, 4)$ and $N(3, 2, -1)$ and perpendicular to the plane $x+y+z-3=0$.
Solution. We will use the equation of the plane passing through the first of the given points:

$$
A(x-2)+B(y+1)+C(z-4)=0
$$

The condition for this plane to pass through the second point and the condition of perpendicularity are given by the equations:

$$
\begin{gathered}
A+3 B-5 C=0 \\
A+B+C=0
\end{gathered}
$$

By eliminating the coefficients $A, B$, and $C$ from the system of equations

$$
\begin{aligned}
& A(x-2)+B(y+1)+C(z-4)=0 \\
& A+3 B-5 C=0 \\
& A+B+C=0
\end{aligned}
$$

we obtain the desired equation in the form:

$$
\left|\begin{array}{ccc}
x-2 & y+1 & z-4 \\
1 & 3 & -5 \\
1 & 1 & 1
\end{array}\right|=0
$$

or

$$
4 x-3 y-z-7=0 \text {. }
$$