b
(b) Since it is an identity in $\lambda $ so satisfied by every value of $\lambda $.
Now put $\lambda = 0 $ in the given equation,
we have $t = \left| {\,\begin{array}{*{20}{c}}0&{ - 1}&3\\1&2&{ - 4}\\{ - 3}&4&0\end{array}\,} \right|\, = - 12 + 30 = 18$.