If the system of equations
$x+2y+3z= 10$
$-x+y+λz=20$
$2x + 3y+λz = 0$
does not possess a unique solution, then a is equal to

$\frac{15}{4}$

The correct answer is Option (2) → $\frac{15}{4}$

Given system

$x+2y+3z=10$

$-x+y+\lambda z=20$

$2x+3y+\lambda z=0$

Coefficient matrix

$A=\begin{pmatrix}1&2&3\\-1&1&\lambda\\2&3&\lambda\end{pmatrix}$

System does not possess unique solution when $|A|=0$

$|A|=\begin{vmatrix}1&2&3\\-1&1&\lambda\\2&3&\lambda\end{vmatrix}$

Expand along first row

$=1\begin{vmatrix}1&\lambda\\3&\lambda\end{vmatrix} -2\begin{vmatrix}-1&\lambda\\2&\lambda\end{vmatrix} +3\begin{vmatrix}-1&1\\2&3\end{vmatrix}$

$=1(\lambda-3\lambda)-2(-\lambda-2\lambda)+3(-3-2)$

$=(-2\lambda)-2(-3\lambda)+3(-5)$

$=-2\lambda+6\lambda-15$

$=4\lambda-15$

Set determinant zero

$4\lambda-15=0$

$\lambda=\frac{15}{4}$

The value of $\lambda$ is $\frac{15}{4}$.