Finding $p$ and $t$ such that the system $px+y+z=1$, $x+2y+4z=t$, $x+4y+10z=t^2$ has one, infinitely-many, or no solutions 
Consider the following equations:
$$\begin{align}
px+\phantom{2}y+\phantom{10}z &=1 \\
x+2y+\phantom{1}4z &=t \\
x+4y+10z &=t^2
\end{align}$$
Now find the values of $p$ and $t$ for which
(a) there is a unique solution
(b) there are infinitely many solutions
(c) there is no solution

My attempt: For (a) we require the determinant $\begin{vmatrix}p&1&1\\1&2&4\\1&4&10\end{vmatrix}\neq 0$. Hence $p\neq 1$.
For (b) we need two of the equations to represent the same line - I tried subbing in the difference between the second and third equation into the first but got nowhere. Similarly for (c).
 A: If $p=1$, your system becomes$$\left\{\begin{array}{l}x+y+z=1\\x+2y+4z=t\\x+4y+10z=t^2.\end{array}\right.$$If you subtract the first equation from both the second one and the third one, you will get$$\left\{\begin{array}{l}x+y+z=1\\y+3z=t-1\\3y+9z=t^2-1.\end{array}\right.$$Since $3y+9z=t^2-1\iff y+3z=\frac{t^2-1}3$, the system has no solution if $t-1\ne\frac{t^2-1}3$ (that is, if $t$ is neither $1$ nor $2$) and it has infinitely many solutions otherwise.
A: System of equations give the below matrix:
$\left(\begin{array}{rrr|r}
p & 1 & 1 & 1\\
1 & 2 & 4 & t\\
1 & 4 & 10 & t^2
\end{array}\right)$
We set third row as $(R_3 - 2 \cdot R_2)$ and second row as $(R_2 - 2 \cdot R_1)$.
$\left(\begin{array}{rrr|r}
p & 1 & 1 & 1\\
1-2p & 0 & 2 & t-2\\
-1 & 0 & 2 & t^2-2t
\end{array}\right)$
If $p = 1$, $t$ must be either $1$ or $2$ for row $2$ and $3$ to be consistent. Also note that it leads to only two equations in $3$ variables.
So we have no solutions, if $p = 1, t \notin (1,2)$
If $p = 1, t \in (1,2)$, we have infinite solutions
If $p \ne 1$, we have unique solution.
A: Thanks to José Carlos Santos who has pointed a flaw in my reasoning.
For the cases (b) and (c), we agree that the range of the system's matrix is a vector space with dimension 2, i.e., a plane. Intuitively, the skew
curve with parametric equations $(x=1,y=t,z=t^2)$ intersects a plane in at most 2 points. Therefore we can have "hope" for the cases where $t$ takes one of these values. (see figure 1).

Fig. 1: Case $p=1$ : The space curve $(1,t,t^2)$ (looking here like a parabola) intersects the 2D range of the LHS of the equation.
let us prove it in an algebraic way.
We know that, necessarily $p=1$.
The system of equations being expressible under the form:
$$xV_1+yV_2+zV_3=V_4$$
More explicitly:
$$x\begin{pmatrix}1\\1\\1\end{pmatrix}+y\begin{pmatrix}1\\2\\4\end{pmatrix}+z\begin{pmatrix}1\\4\\10\end{pmatrix}=\begin{pmatrix}1\\t\\t^2\end{pmatrix}\tag{0}$$
As the 3 columns on the LHS belong to the range of the system's matrix and are dependent (the range has dimension 2), we can take a simple basis of this range, out of which we can "sweep" all this range:
System (0) can be written under the form:
$$u\underbrace{\begin{pmatrix} \ \ 1\\ \ \ 0\\-2\end{pmatrix}}_U+v\underbrace{\begin{pmatrix}0\\1\\3\end{pmatrix}}_V=\begin{pmatrix}1\\t\\t^2\end{pmatrix}$$
Before going further, check that
$$U+V=V_1, \ \ U+2V=V_2, \ \ U+4V=V_3$$
$$\begin{cases}u&=&1\\v&=&t\\-2u+3v&=&t^2\end{cases}\tag{1}$$
which will be fulfilled iff:
$$-2+3t=t^2\tag{2}$$
i.e., iff $t$ is a root of the quadratic equation (2),
$$t=1 \ \ \text{or} \ \ t=2.$$
In these two cases, there exists an infinite number of solutions of systems obtained from (0) by replacing $t$ by $1$ or by $2$:
$$\text{case} \ t=1 : \ \begin{cases}
x+\phantom{2}y+\phantom{10}z &=1 \\
x+2y+\phantom{1}4z &=1 \\
x+4y+10z &=1
\end{cases} \text{and }  \text{case} \ t=2 : \ \begin{cases}
x+\phantom{2}y+\phantom{10}z &=1 \\
x+2y+\phantom{1}4z &=2 \\
x+4y+10z &=4
\end{cases}$$
Indeed, in both systems, the 3rd line $L_3$ can be cancelled, being a combination of the 2 others $L_3=-2L_1+3L_2$, giving the intersection  of two affine planes, i.e., an affine line of solutions in both cases:
$$\text{case} \ t=1 : \ \begin{cases}
x&=&1+2a \\ y&=&-3a \\ z&=&a
\end{cases} , \ \  \text{cases} \ t=2 : \ \begin{cases}
x&=&2a \\ y&=&1-3a \\ z&=&a
\end{cases}$$
Otherwise, if $t\ne 1,2$ there will be no solution.
