`"Let "plambda^(4) + qlambda^(3) +rlambda^(2) + slambda +t =|{:(lambda^(2)+3lambda,lambda-1, lambda+3),(lambda+1, -2lambda, lambda-4),(lambda-3, lambda+4, 3lambda):}|` be an identity in `lambda`, where p,q,r,s and t are constants. Then, the value of t is..... .

Explanation

To solve the problem, we need to evaluate the determinant given in the question and equate it to the polynomial expression on the left side. We will find the value of \( t \) by expanding the determinant.

### Step 1: Write down the determinant
The determinant we need to evaluate is:
\[
D = \begin{vmatrix}
\lambda^2 + 3\lambda & \lambda - 1 & \lambda + 3 \\
\lambda + 1 & -2\lambda & \lambda - 4 \\
...