Determine the values of $a$ for which the following linear system has (i) no solutions, (ii) infinitely many solutions, (iii) exactly one solution:
\begin{cases} x + 2y − 3z = 4 \\[4px] 3x − y + 5z = 2 \\[4px] 4x + y + (a^2 − 14)z = a + 2 \end{cases}
On
Without knowing what tools you have available, it is difficult to give an answer as expected by your instructor. However, a standard procedure is Gaussian elimination on the system's matrix. \begin{align} \left[\begin{array}{ccc|c} 1 & 2 & -3 & 4 \\ 3 & -1 & 5 & 2 \\ 4 & 1 & a^2-14 & a+2 \end{array}\right] &\to \left[\begin{array}{ccc|c} 1 & 2 & -3 & 4 \\ 0 & -7 & 14 & -10 \\ 0 & -7 & a^2-2 & a-10 \end{array}\right] &&\begin{aligned}R_2&\gets R_2-3R_1\\ R_3&\gets R_3-4R_1\end{aligned} \\ &\to \left[\begin{array}{ccc|c} 1 & 2 & -3 & 4 \\ 0 & -7 & 14 & -10 \\ 0 & 0 & a^2-16 & a \end{array}\right] &&R_3\gets R_3-R_2 \end{align} You can clearly see that the transformed system has a unique solution if and only if $a^2-16\ne0$.
For $a^2-16=0$, the third row has a nonzero entry in the last column, which says the system has no solution.
As the linear system is a non-homogeneous one (why?), it has one unique solution iff the matrix of coefficients of the system is regular iff its determinant is non-zero, so:
$$\begin{vmatrix} 1&2&\!-3\\ 3&\!-1&5\\ 4&1&a^2-14\end{vmatrix}\stackrel{R_2-3R_1,\,R_4-4R_1}\longrightarrow\begin{vmatrix} 1&2&\!-3\\ 0&\!-7&14\\ 0&\!-7&a^2-2\end{vmatrix}=-7(a^2-2)+98=0\iff$$$${}$$
$$\iff\;\;a^2=100\iff a=\ldots$$$${}$$
So now you know when the system has one unique solution. For the two values of $\;a\;$ for which there is no unique solution, you have to check each one and find out whether you get infinite solutions (as I think you are working over the real numbers), or whether you get no solution at all...and you can use the above calculations...