I have solved most of the exercises below, except for the last one. Are the solutions correct? What to do for the last one?
Given three points $a = (1,4,0), b=(2,1,5), c=(3,5,2)$ in $\mathbb{R^3}$, find each of the following
$i)$ vector from $a$ to $b$
My solution: $$\vec{ab} = b - a$$ $$= (2,1,5)-(1,4,0)$$ $$= (1,-3,5)$$
$ii)$ parametric form of the equation of the line through $a$ and $b$
My solution: $(x,y,z) = t\times \vec{ab} = t(1,-3,5)$ $$x=t, y=-3t, z=5t $$
$iii)$ A parametric equation of the plane containing $a$, $b$ and $c$
My solution: $$(x,y,z) = (1,4,0) + t_{1}(2,1,5) + t_{2}(3,5,2)$$ $$x=1+2t_{1}+3t_{2}, \space\space y=4+t_{1}+5t_{2}, \space\space y=0+5t_{1}+2t_{2}$$
$iv)$ A Cartesian equation of the plain containing $a$, $b$ and $c$
My solution: I need a vector and a plane. Use $\vec{ab} = (1,-3,5)$ $$v = A(x-x_{0})+B(y-y_{1})+C(z-z_{0})$$ with perpendicular line
$v)$ A vector perpendicular to the plane containing $a$, $b$ and $c$
Your solution to $(i)$ is correct.
For $(ii)$, you have a parametrization of a line in the direction of $\vec{ab}$, but not going through $\vec {ab}$. You can see this by trying to solve $a = (1,4,0) = t(1,-3,5)$. If your line does go through $a$ and $b$ there should be a solution to that equation, but it's clear that there is not.
What you need to do is just shift your line so that it goes through $a$ and $b$. The easiest way to do this is to add $a$ OR $b$ (whichever you prefer) to the equation you've already derived. That is, one solution to this problem will by $(x,y,z) = t(1,-3,5) + (1,4,0)$. You can check that this line does indeed pass through both $a$ and $b$.
For $(iii)$, Your plane will be of the form $$\text{(vector in the plane)} + t_1\text{(vector parallel to the plane)}+t_2\text{(another vector parallel to the plane}) = (x,y,z)$$
So notice that $b$ and $c$ are NOT necessarily parallel to the plane, but $\vec {ab}=\vec b - \vec a$ and $\vec {ac} = \vec c - \vec a$ (or any other combination of $\{\vec {ab},\vec{ac}, \vec{bc}\}$ provided you show that the two you choose are not parallel to each other) are. And again, you have $3$ choices for the vector in the plane.
$(iv)$ Here you need to find a normal vector $(n_1, n_2, n_3)$. Then your plane will have the equation $(n_1, n_2, n_3) \cdot (x-x_0, y-y_0, z-z_0) = n_1(x-x_0)+n_2(y-y_0)+n_3(z-z_0) = 0$.
To find a normal vector, you can use the cross product. First you'll need $2$ non-parallel vectors, which are each parallel to your plane. Because you have $3$ points, you can just use $\vec {ab}$ and $\vec {ac}$ (verify first that these two vectors are not collinear). Then take their cross product to get the normal vector $(n_1, n_2, n_3)$. All that's left to find at this point is a vector in the plane $(x_0, y_0, z_0)$, luckily the problem has given you $3$, so just choose your favorite, plug it in, and you have your answer to this one.
$(v)$ I showed you how to find this in $(iv)$ with the cross product.
But here's another way:
Notice that what you want is a vector which is orthogonal to $\vec {ab}$, $\vec {ac}$, and $\vec {bc}$. This means that you need to solve the equations: $\begin{cases} (\vec b - \vec a) \cdot \vec n = 0 \\ (\vec c - \vec a) \cdot \vec n = 0 \\ (\vec c - \vec b) \cdot \vec n = 0 \end{cases}$. If you let $\vec {ab} = (a_1, a_2, a_3), \vec {ac} = (b_1, b_2, b_3), \vec {bc} = (c_1, c_2, c_3),$ and $\vec n = (n_1, n_2, n_3)$, then notice that this is equivalent to the set of equations: $\begin{cases} a_1n_1 + a_2n_2 + a_3n_3 = 0 \\ b_1n_1 + b_2n_2 + b_3n_3 = 0 \\ c_1n_1 + c_2n_2 + c_3n_3 = 0 \end{cases}$ which is exactly the same as the matrix equation: $$\begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix} \begin{bmatrix} n_1 \\ n_2 \\ n_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$ Now just solve for $(n_1, n_2, n_3)$.