Find equation of the plane passing through the line $x+y=2, y-z=3$ and is perpendicular to the plane $2x+3y+4z=5$

Solution 1:

The line is $~x = t, y = - t + 2, z = - t - 1$ and the direction vector of the line is $\vec d = (1, -1, -1)$ and not $(1, -1, 1)$. That is your first mistake. The desired plane is also perpendicular to plane $2x + 3y + 4z = 5$ with normal vector $\vec n = (2, 3, 4)$. Cross product of the direction vector of the line contained in the plane and the normal vector of the plane perpendicular to the desired plane would give us the normal vector of the desired plane.

$\vec n \times \vec d = (1, 6, -5)$ so the equation of the plane is $x + 6y - 5z = k$

As the line $x + y = 2, y - z = 3$ is in the plane,

$x + 6y - 5z = x + y + 5 (y - z) = 2 + 5 \cdot 3 = 17$

So the equation of the desired plane is $x + 6y - 5z = 17$

The book solution simply uses the fact that the equation of the given line should satisfy the equation of the plane as the line is contained in the plane. As the line is $x + y -2 = 0, y - z - 3 = 0$, any plane with the equation $(x + y - 2) + \lambda (y - z - 3) = 0$ would contain the given line.