If the system of linear equations
$$x + ky + 3z = 0$$
$$3x + ky - 2z = 0$$
$$2x + 4y - 3z = 0$$
has a non-zero solution (x, y, z), then $$\frac{xz}{y^2}$$ is equal to:

For a homogeneous system to possess a non-zero solution, the determinant of its coefficient matrix must be zero. We therefore begin by writing the determinant $$\Delta$$ of the coefficients

$$ \Delta=\begin{vmatrix} 1 & k & 3\\ 3 & k & -2\\ 2 & 4 & -3 \end{vmatrix}. $$

We expand this determinant along the first row, stating the expansion formula: for a 3 × 3 determinant $$\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix}=a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{33}-a_{23}a_{31})+a_{13}(a_{21}a_{32}-a_{22}a_{31}).$$ Applying it here we get

$$ \Delta =1\big(k(-3)-(-2)(4)\big) -k\big(3(-3)-(-2)(2)\big) +3\big(3\cdot4-k\cdot2\big). $$

We simplify term by term. First term:

$$k(-3)-(-2)(4)=-3k+8.$$

Second term:

$$3(-3)-(-2)(2)=-9+4=-5,$$ so

$$-k(-5)=+5k.$$

Third term:

$$3\cdot4-k\cdot2=12-2k,$$ hence

$$3(12-2k)=36-6k.$$

Adding all three contributions,

$$ \Delta=(-3k+8)+5k+(36-6k)=44-4k. $$

For a non-trivial solution we set $$\Delta=0,$$ giving

$$44-4k=0\;\;\Longrightarrow\;\;k=11.$$

Now we substitute $$k=11$$ into the original equations:

$$ \begin{aligned} x+11y+3z&=0\quad&(1)\\ 3x+11y-2z&=0\quad&(2)\\ 2x+4y-3z&=0\quad&(3) \end{aligned} $$

From equation (1) we isolate $$x$$:

$$x=-11y-3z. \quad -(4)$$

We substitute this value of $$x$$ into equation (2):

$$ 3(-11y-3z)+11y-2z=0. $$

Expanding we have

$$-33y-9z+11y-2z=0,$$ so

$$-22y-11z=0.$$ Dividing by $$-11$$ gives

$$-2y-z=0\;\;\Longrightarrow\;\;z=-2y. \quad -(5)$$

Substituting the value of $$z$$ from (5) into equation (4) we obtain

$$ x=-11y-3(-2y)=-11y+6y=-5y. \quad -(6) $$

Thus a typical non-zero solution is proportional to

$$ (x,y,z)=(-5y,\,y,\,-2y),\qquad y\neq0. $$

We need the value of $$\dfrac{xz}{y^{2}}$$. Using the expressions just found,

$$ xz=(-5y)(-2y)=10y^{2}, $$ and therefore

$$ \frac{xz}{y^{2}}=\frac{10y^{2}}{y^{2}}=10. $$

Hence, the correct answer is Option 3.