Planar geometry question HELP PLEASE!?

Question

The line L through the origin is normal to the plane: 2x-y-z=4. Find the point in which L meets the plane x+y-2z=2.

Answer 1

♠ thus unit vector normal to the plane 2x-y-z=4 is
n= (2, -1, -1) / √(2^2 +(-1)^2 +(-1)^2) = (2/√6, -1/√6, -1/√6);
♣ thence the equation of the line L in parametric form is vector
r= (x,y,z) =n*p;
♦ plugging r into the second plane x+y-2z=2 we get:
2p/√6 -p/√6 +2p/√6=2, hence p=2*√6 /(2 -1+2) = (2/3)*√6;
the point in question is
r =((2/3)*√6 *2/√6, (2/3)*√6*(-1)/√6, (2/3)*√6*(-1)/√6)=
= (4/3, -2/3, -2/3);

Answer 2

Add the two expressions
2x -y -z =4
+
x + y -2z =2
----------------
3x -3z = 6 divide the whole equation by 3

x -z = 2

therefore x = z + 2

since x is ( z + 2) substitute this value in the equations and u will get the modified equations as :

2(z+2) -y -z =4-------------> First expression
and
z+2 + y -2z =2--------------> Second expression

Now simplify the 1st eqation:

2z +4 - y -z =4

z -y = 0
z = y -----------------------> Simplified First expression

Simplify the second equation :

y -z = 0
y = z-----------------------> Simplified Second expression

Thus the conclusion is the point where the y and z axis meet is the point where L meets the plane.