ABCD is a parallelogram. The position vector of the points A,B and C are respectively, 4i + 5j -10k , 2i-3j+4k and -i+2j+k. Find the vector equation of the line BD. Also reduce it to cartesian form.

A=(4,5,-10); B=(2,-3,4), C=(-1,2,1), Let D=(x,y,z)
In a parallelogram diagonals bisect each other.
So, mid point of AC = mid point of BD
$$\begin{gathered} \left(\frac{4-1}{2},\frac{5+2}{2},\frac{-10+1}{2}\right)=\left(\frac{2+x}{2},\frac{-3+y}{2},\frac{4+z}{2}\right); \\ \left(\frac{3}{2},\frac{7}{2},\frac{-9}{2}\right)=\left(\frac{2+x}{2},\frac{-3+y}{2},\frac{4+z}{2}\right); \\ 2+x=3; -3+y=7; 4+z=-9; \\ x=1; y=10; z= -13; \end{gathered}$$
Now B=(2,-3,4)=(x1,y1,z1) and D= (1,10,-13)=(x2,y2,z2)
Dr's of BD = (x2,x1, y2-y1, z2-z1)=(-1, 13,-17)

Equation of BD in vector form is,
(x1,y1,z1)+t(a,b,c)=(2,-3,4)+t(-1, 13,-17)

Its cartesian form would be,
$\frac{x-2}{-1}=\frac{y+3}{13}=\frac{z-4}{-17}$