How to solve a few linear algebra vector questions

Hi, I have a few questions I need help in linear algebra involving vectors.

1. Given a vector equation r(t) = (3-4t)i+(0+4t)j+(-2+2t)k , rewrite this in terms of the symmetric equations for the line.

(quotient involving x) = ?
(quotient involving y) = ?
(quotient involving z) = ?

2. Consider the planes 3x+4y+5z=1 and 3x+5z =0
a) Find a unit vector U with positive first coordinate that is parallel to both planes. (i,j,k)
b) Using part A and the unique point P on the y-axis which is on both planes being (0, (1/4), 0), find a vector equation for the line of intersection of the two planes (i,j,k)

3. Find the distance from the point (3, 1, 1) to the line x=0,y=1+4t,z=1+3t


I would deeply appreciate any help in at least one of the questions.
Thank you

1. A line can be written as:
r(t) = x(t) i + y(t) j + z(t) k
r(t) = (3 - 4t) i + (4t) j + (2t - 2) k

So break it into the components:
x(t) = 3 - 4t
y(t) = 4t
z(t) = 2t - 2

Symmetric equations:
t = (3 - x)/4 = y/4 = (z + 2)/2
==========================
2. Find the norms of the planes:
n1 = <3,4,5>
n2 = <3,0,5>

The normal vectors must be perpendicular to u for it to be in the plane:
u • n1 = 0
• <3,4,5> = 0
3u1 + 4u2 + 5u3 = 0

u • n2 = 0
• <3,0,5> = 0
3u1 + 5u3 = 0

Subtract the equations to find:
u2 = 0.

Find u (unit vector) by dividing by the length:
u = <3,0,5> / √(3² + 5²)
u = <3,0,5> / √34
==========================
3. Shortest distance occurs when the vector from the point to the line is perpendicular to the line
• r'(t) = 0
<(x - 3), (y - 1), (z - 1)> • <0, 4, 3> = 0
4(y - 1) + 3(z - 1) = 0
4y + 3z = 7
Replace the x,y,z with the equations for the line:
4(1 + 4t) + 3(1 + 3t) = 7
4 + 16t + 3 + 9t = 7
25t = 0
Therefore, the minimum distance occurs when t = 0
t = 0

Find the point where r(t) is at t = 0:
r(t) = <0, (1 + 4t), (1 + 3t)>
r(0) = <0, 1, 1>

Find the distance:
D = ||
D = √((0 - 3)² + (1 - 1)² + (1 - 1)²)
D = 3