Problem 6. Find the equation of the plane $ABF$, if $F(-4; 8; -3), A(-1; 2; -3), B(3; 4; 1)$.
Solution. If $M(x, y, z)$ is an arbitrary point on the plane $A B F$, then $\overline{B M}=k B A+n B F$ or $(x-3 ; y-4; z-1)=k(-4 ;-2 ;-4)+n(-7 ; 4 ;-4)$.

From this, we obtain the system of equations:

$$
\left\{\begin{array}{l}
x-3=-4 k-7 n \\
y-4=-2 k+4 n \\
z-1=-4 k-4 n
\end{array}\right.
$$

Multiply both sides of the second equation by -2. After that, add equation (2) term by term to equations (1) and (3). We get

$$
\begin{aligned}
& x-3-2(y-4)=-15 n \\
& z-1-2(y-4)=-12 n
\end{aligned}
$$

Multiply equation (4) by 4, equation (5) by -5. After term-by-term addition of these equations, we get the answer: $5 z-2 y-4 x+15=0$.