In general, the equation of the plane normal to $\bf n$ containing $(x_0,y_0,z_0)$
is
$$
{\bf n}\cdot[ (x-x_0) i +(y-y_0) j + (z-z_0)k]=0.\qquad\qquad(1)
$$
In your problem, we need to find a normal vector for the plane.
To do this, we need two independent vectors in the plane. The direction vector of the line, $-i+j+k$, will serve as one. For the other, take the vector formed by the difference of $A$ and any point on the line. Since the line is in the plane $B=2i+3j+4k$ will serve.
Now $AB= 4i +2j -k$ is your second vector in the plane.
To obtain the vector normal to the plane, take the cross product of $AB$ and the direction vector of the line.
$${\bf n}=AB\times( -i+j+k )=\Biggl|\matrix{ i&j&k\cr 4&2&-1\cr -1&1&1\cr}\Biggr|=
3 i - 3j +6k.
$$
The equation of the plane is, using (1) with $B$ as the point:
$$
{\bf n}\cdot[ (x-2) i +(y-3) j + (z-4)k]=0.
$$
To see that this gives the same result of the author:
$$
\eqalign{
&{\bf n}\cdot[ (x-2) i +(y-3) j + (z-4)k]=0 \cr
\iff&(3i-3j+6k)\cdot[ (x-2) i +(y-3) j + (z-4)k]=0 \cr
\iff&(3i-3j+6k)\cdot[ (x i +yj + zk) +( -2i-3j-4k)]=0 \cr
\iff&(3i-3j+6k)\cdot (x i +yj + zk) =(3i-3j+6k)\cdot( 2i+3j+4k) \cr
\iff&(3i-3j+6k)\cdot (x i +yj + zk) =21 \cr
\iff&(i-j+2k)\cdot (x i +yj + zk) =7 \cr}
$$
Final edit:
More simply, the author is using the vector plane equation
$$\tag{2}{\bf n}\cdot {\bf r} ={\bf a}\cdot{\bf n}$$ where $\bf n$ is the normal vector, $\bf r$ is the position vector to the plane and
$\bf a$ is the position vector of a point on the plane. So, once you find $\bf n$, take ${\bf a}=2i+3j+4k$, and substitute into (2).