Finding equations of a plane/line/projection of a point

1. Find an equation of a plane passing through the point P(-3, 1, 0) and parallel to the lines
L : x=4t, y=1+t, z=2-3t
L : x=-2+t, y=-1-2t, z=3+2t, t€R

2. Find an equation of a line passing through the point P(1, 0, 5) and parallel to the line
l : x+2y-z+3=0,3x+y+2z+4=0

3. Find a projection of the point P(2, -3, 1) on the plane -x+3y-2z-1=0. Find the distance from the point to the plane.

THANK YOU IN ADVANCE :)

1)
The plane's normal vector will be perpendicular to the direction vectors of the lines.
So we take the cross product of the two lines' direction vectors to get the plane's normal vector.
<4, 1, -3> X <1, -2, 2>
I get
i(-4) + j(-11) + k(-9) or, equivalently, 4i + 11j + 9k
so the plane will be 4x + 11y + 9z = C where C is a constant.
To find C, plug in the point (-3, 1, 0):
4(-3) + 11(1) + 9(0) = C = -1

so the plane is 4x + 11y + 9z = -1

2)
The line will be perpendicular to the normal vectors of the planes:
x+2y-z+3=0 and 3x+y+2z+4=0
The normal vectors are <1, 2, -1> and <3, 1, 2>
so we take their cross product to find the line's direction vector.
<1, 2, -1> X <3, 1, 2>
I get
i(5) + j(-5) + k(-5), or, equivalently, i - j - k.
So the line's vector equation is (1, 0, 5) + t<1, -1, -1>
Then the line's parametric equations are:
x = 1 + t
y = -t
z = 5 - t

3)
The projection of the point onto the plane can be found by drawing a line perpendicular to the plane, connecting the point and the plane. The point where this line crosses the given plane will be the projection of the point onto the plane.
The equation of such a line is
(2, -3, 1) + <-1, 3, -2>t
The (2, -3, 1) is given as the point and the direction vector <-1, 3, -2> comes from the plane's normal vector.

The parametric form of the equation above is
x = 2 - t
y = -3 + 3t
z = 1 - 2t

to find the point where this line crosses the plane, i.e. the point we're looking for, we simply plug in x, y, and z into the plane equation and solve for t.

So we have:
-(2-t) + 3(-3+3t) - 2(1-2t) - 1 = 0
-2 + t - 9 + 9t - 2 + 4t - 1 = 0
-14 + 14t = 0
14t = 14
t = 1

So the point we're looking for can be found by plugging t=1 into the parametric equation of the line we've found.
x = 2 - t
y = -3 + 3t
z = 1 - 2t

x = 2 - 1
y = -3 + 3(1)
z = 1 - 2(1)

x = 1
y = 0
z = -1

so the projection is (1, 0, -1) and the distance is given by the distance formula from (2,-3,1) to (1,0,-1):
sqrt((2-1)^2 + (-3-0)^2 + (1 - -1)^2)
sqrt(1^2 + (-3)^2 + 2^2)
sqrt(1+9+4)
sqrt(14)

Hope this helps.