Answer: The coefficient matrix is \(A=\begin{pmatrix}14&-4&6\\1&1&k\\-21&6&-9\end{pmatrix}\).

Its determinant is

\(\det(A)=14\begin{vmatrix}1&k\\6&-9\end{vmatrix}+4\begin{vmatrix}1&k\\-21&-9\end{vmatrix}+6\begin{vmatrix}1&1\\-21&6\end{vmatrix}\)

\(=14(-9-6k)+4(-9+21k)+6(6+21)=0\).

So the system does not have a unique solution for any value of \(k\).

Geometrically, the first and third equations represent two distinct parallel planes, since \((-21,6,-9)=-\tfrac32(14,-4,6)\) but \(14\neq -\tfrac32\cdot 5\).

The second plane is not parallel to these, since \((1,1,k)\) is not a scalar multiple of \((14,-4,6)\).

A system of three linear equations in \(x,y,z\) has a unique solution only if the determinant of its coefficient matrix is non-zero.

Here the coefficient matrix is

\(A=\begin{pmatrix}14&-4&6\\1&1&k\\-21&6&-9\end{pmatrix}.\)

Expand \(\det(A)\) along the first row:

\(\det(A)=14\begin{vmatrix}1&k\\6&-9\end{vmatrix}-(-4)\begin{vmatrix}1&k\\-21&-9\end{vmatrix}+6\begin{vmatrix}1&1\\-21&6\end{vmatrix}.\)

So

\(\det(A)=14(-9-6k)+4(-9+21k)+6(6+21).\)

Hence

\(\det(A)=(-126-84k)+(-36+84k)+162=0.\)

Therefore \(\det(A)=0\) for all values of \(k\), so the system cannot have a unique solution.

Geometrically, each equation represents a plane.

The first and third equations are

\(14x-4y+6z=5\)

and

\(-21x+6y-9z=14.\)

Their normal vectors are \((14,-4,6)\) and \((-21,6,-9)\), and

\((-21,6,-9)=-\tfrac32(14,-4,6),\)

so these two planes are parallel.

They are not the same plane, because the constants are not in the same ratio:

\(-\tfrac32\cdot 5=-\tfrac{15}{2}\neq 14.\)

So the first and third planes are distinct parallel planes.

The second plane has normal vector \((1,1,k)\), which is not a scalar multiple of \((14,-4,6)\), so it is not parallel to those two planes.

Thus the geometry is: two distinct parallel planes, and a third plane that is not parallel to them.