Vectors - Lines and Planes

1. The line passing through a point $a$ and parallel to a line $r=b+ct$ is $r=a+ct$, so we get the parametric equation $$ r=5i-4j+2k +(5i-2j+k)t=(5+5t)i+(-4-2t)j+(2+t)k\\=xi+yj+zk\\\implies x=5+5t,\;y=-4-2t,\; z=2+t\\\implies t={x-5\over 5}={y+4\over -2}=z-2$$which is the symmetric equation.

2. The plane passing through the point $a$ and perpendicular to the line $r=b+ct$ is $(r-a)\cdot c= 0$ where $r=(x,y,z)$. So we get $$ (x-5,y+4,z-2)\cdot (5,-2,1)=0\\ \implies 5(x-5)-2(y+4)+z-2=0 \\ \implies 5x-2y+z=25+8+2=35$$

3. The plane containing two parallel lines $r=a+ct$ and $r=b+ct$  contains also the line passing through the points $a,b$ which is $r=a+(b-a)t$. So its normal vector is perpendicular to both vectors $c, b-a$. So we can take the normal vector to be $n=c\times (b-a)$ and the quation for the plane is $(r-a)\cdot n=0$. Now we have $b-a=(5,-4,2)-(1,-1,0)=(4,-3,2)$ and $$ n=(4i-3j+2k)\times (5i - 2j + k) = i + 6j+7k$$So th equation of the plane is $$ (x-5,y+4,z-2)\cdot (1,6,7)=0 \\ \implies x-5+6y+24+7z-14=0 \\ \implies x+6y+7z=-5$$