So Im going to give a correction for the last exercice. Humbert has already revised its factorisation and is perhaps waiting impatiently to check his results. I did this one under notepad to copy it directly, but I couldn't do the last part which is difficult.

1)

a) The vector AB(2;3;2) is a direction vector of Δ.

Then,

x = 8+2t

y = 3t t=IR

z = 8+2t

b)2 lines are coplanar if they are not parallel, nor secant.

let's take the vector u(3,2,-2) a direction vector of (D).

-lets prove they are not parallel:

AB(2,3,2) and u(3,2,-2) are not collinear (there is no real number k such as AB=ku)

then (Delta) and (D) are not parallel.

-lets prove they are not secant:

We have to resolve the system to eventually find a point of intersection:

x = - 5 + 3 s

y = 1 + 2 s

z = - 2 s

x = 8+2t

y = 3t

z = 8+2t

<=>

-5+3s = -2s <=> s= 1

y = 1 + 2 s

z = - 2 s

x = z

y = 3t

z = 8+2t

<=>

s = 1

y = 3

z = -2

y = 3 <=> t = 1

z = 10

The system has no solution then the lines (AB) and Δ are not secant.

The lines (AB) and Δ are not parallel nor secant, they are not coplanar.

2)

a) The plane (P) is parallel to (D) and contains Δ. The vector n (2 ; - 2 ; 1) is normal to (P).

Then the vectors AB(2,3,2) and u(3,2,-2) are direction vectors of the plane (P).

Let's do The scalar products to check if n is normal to (P):

n • u = 3 * 2 + 2 * (– 2) + (– 2) * 1 = 6 – 4 – 2 = 0

AB • n = 2 * 2 + 3 * (– 2) + 2 * 1 = 4 – 6 + 2 = 0

n is orthogonal to u and AB, and therefore orthogonal to (P), n is a normal vector to (P).

Knowing the normal vector, the cartesian equation of (P) has the form: 2x-2y+z+d=0

The plane (P) contains Delta and the point A(8,0,8) then

2*8-2*0+8+d=0 <=> d=-24

the cartesian equation of (P) is 2x-2y+z-24=0

b)lets take the point M(x,y,z).

we know the coordinates of M verify the system:

x = - 5 + 3 s

y = 1 + 2 s

z = - 2 s

Let's calculte the distance of M to (P) with the known formula:

d(M,(P))=|axM+byM+czM|/sqrt(a²+b²+c²)

then

d(M,(P))= |2x-2y+z-24|/sqrt(2²+(-2)²+1)=|2(-5+3s)-2(1+2s)-2s|/sqrt(9)=|-36|/3=12

We find a real number independent of the point M. Then the distance of any point M of (D)

to (P) is independent of M. The interpretation of this result is obvious since (D) is parallel to (P).

c) To give a parametric equation of (P) with (xOy) let's resolve the system:

2x-2y+z-24=0

z=0

<=>

2x=2y+24

z=0

<=>

x=y+12

z=0

The system of parametic equations is:

x=t+12

y=t t=IR

z=0

3)