Show that the lines 5−x=y=z+3 and x=2y=z are coplanar.

Given lines are
5−x=y=z+3 and x=2y=z
5−x=y=z+3 — (1)
x=2y=z — (2)
Let the equation of plane passing through (5,0,-3) be
a(x-5)+by+c(z+3)=0 — (3)
Normal to the plane 3 is ⊥ to (1) — (4)
Normal to the plane 3 is ⊥ to (2) — (5)
On solving (4) and (5), we get
-a+b+c=0
a-2b+c=0
Solving these equations we get a=c and b=0
Therefore the equation of the plane is x-z-2=0
Let us check whether the point (0,0,0) lies on the plane x-z-2=0. The point (0,0,0) does not satisfy this plane. The point (5,0,-3) is on the line 5-x=y=z+3, so we use it to find the equation of the plane.
Let the equation of the plane be a(x-5)+by+c(z+3)=0. The direction ratios of the first line are <1,1,1> and the second line are <1,1/2,1>.
Therefore we have:
a-b-c = 0
a-b/2-c=0
Solving these we get b=0 and a=c.
Putting this in the equation of the plane we get x-z-2=0.
Let us check whether the point (0,0,0) lies on (6)
0-0-2=0 which is false. Hence the equation of the plane is incorrect.
Given lines are
5−x=y=z+3 — (1)
x=2y=z — (2)
Let the equation of plane passing through (5,0,-3) be
a(x−5)+by+c(z+3)=0 — (3)
Normal to the plane 3 is ⊥ to (1) — (4)
Normal to the plane 3 is ⊥ to (2) — (5)
On solving (4) and (5), we get
−a+b+c=0
a−2b+c=0
Solving these equations, we get a=c and b=0
Therefore, the equation of the plane is x−z−2=0
Let us check whether point (1,2,1) lies on the line x=2y=z. It does.
1-1-2=-2≠0. This point is not on this plane. There must be an error in the solution.
Let's use a different method. From (1), we can write x = 5-k, y = k, z = k-3 for some parameter k. From (2), we can write x = m, y = m/2, z = m for some parameter m.
If the lines are coplanar, then we should be able to find a plane that contains both lines. Let's try to find the equation of the plane containing the points (5,0,-3), (4,1,-2), and (1,1/2,1) from the respective lines.
Using the formula for the equation of a plane through three points, we can find the equation of the plane which turns out to be 17x-7y-64z-102=0.
Now let's check if the other points of the line also lie on this plane.
This method appears to be quite cumbersome and prone to calculation errors. A more concise approach should be used.