The value of $\lambda$ do the planes $x-y+z+1=0,\lambda x+3y+2z-3=0,3x+\lambda y+z-2=0$ from a triangular prism must be

$\therefore {\mathrm{\Delta }}_{ }=\left|\begin{array}{ccc}1& -1& 1\\ \lambda & 3& 2\\ 3& \lambda & 1\end{array}\right|=(\lambda -4)(\lambda +3)$              (i)

Also, ${\mathrm{\Delta }}_1=\left|\begin{array}{ccc}-1& 1& 1\\ 3& 2& -3\\ \lambda & 1& -2\end{array}\right|=5(\lambda -2)$                 (ii)

${\mathrm{\Delta }}_2=\left|\begin{array}{ccc}1& 1& 1\\ \lambda & 2& -3\\ 3& 1& -2\end{array}\right|=(3\lambda -16)$                        (iii)

${\mathrm{\Delta }}_3=\left|\begin{array}{ccc}1& -1& 1\\ \lambda & 3& -3\\ 3& \lambda & -2\end{array}\right|=(\lambda +3)(\lambda -2)$             (iv)

the given planes form a triangular prism, then we know ${∆}_{ }=0$ and none of ${\mathrm{\Delta }}_1,{\mathrm{\Delta }}_2,{\mathrm{\Delta }}_3$ is zero Here from Eqs. (i), (ii), (iii) and (iv: we find that if $\lambda =4,\text{ then }{\mathrm{\Delta }}_{ }=0\text{ and none of }{\mathrm{\Delta }}_1,{\mathrm{\Delta }}_2,{\mathrm{\Delta }}_3$ is zero.