Finding pyramid height equation. Given 4 vector points. Okay, I am having very hard time with this...Gave up.
A(3;4;5), B(1;2;1), C(-2;-3;6), D(3;-6;-3) 
I completed a), asked me to find the area of ABC triangle, got 5*sqrt(14)
b) asked me to find ABC plane equation, used matrix, got ABC = -30x + 22y + 4z - 18
c) asked me to find the equation of AD...used (x-3)/-5=(y-4)/-7=(z-5)/1, transformed it into matrix, and got AD = x + z * 9/5 + 2/5
d) Asks me to find equation of DD1( = h, height).
e) Asks me to find coordinates of DD1
f) Height length 
g) Volume
Hope someone can help me with this. Thank you in advance. 
Added:
THE PYRAMID
 A: For the equation of the line that contains the height, you have a point in it (D) and the normal vector of the base gives the direction of the line, so everything is done.
For the coordinates of the height vector, you could find the intersection of the line whose equation we've just calculates and the base, which gives the point $D_1$. Now we now both endpoints of the vector, so it's easy to find its coordinates.
Finally, we can find the height lenght $h$ by calculating the modulus of the height vector, and the area $S$ of the base using the cross product. Now the volume equals
$$V=\frac{Sh}{3}$$
A: $D D_1$ is perpendicular to the plane of $ABC$ - so parallel to the normal vector $(-30,22,4)$ it also goes through $D$
$$D D_1(t) = (3,-6,-3)+t(-30,22,4) = (3-30t,-6+22t,-3+4t)
$$
To find co-ordinates of $D_1$ solve for the value of $t$ where the line intersects the plane
$$ -30(3-30t)+22(-6+22t)+4(-3+4t)-18=0 $$
then plug that value of $t$ into the equation of the line $D D_1(t)$to get the co-ordinates $D_1$
