Yup its correct, thank u!

May i know how you get the eqn of the line for the second line?

For the second line OB, they said line OB. For a line, it must consist of a point and direction vector which in this case is the origin so (0,0,0) and the direction vector is simply OB direction which is given

Is this the one you’re asking ?^

Yup but i was wondering where the (2,1,2) came from cos OB is (1,3,2) for second eqn

He's drawing a line from B to the plane such that this line is perpendicular to the plane.

The equation of the line has the fixed point B. The direction vector must be perpendicular to the plane so he's just using the plane's normal vector (2 1 2)

The point of intersection of this line and the plane will satisfy the equations of both.

Once this point is found, the vector from B to this point (which lies on the line, so this vector is perpendicular to the plane) can be found by subtracting OB from the position vector of this point. Its magnitude is simply the perpendicular distance from B to the point

Ohhhh ok!! I get it, thank u all so much!

Welcome