`"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..... .

Answer

To find the value of \( t \) in the identity

\[ p\lambda^4 + q\lambda^3 + r\lambda^2 + s\lambda + t = \left| \begin{array}{ccc} \lambda^2 + 3\lambda & \lambda - 1 & \lambda + 3 \\ \lambda + 1 & -2\lambda & \lambda - 4 \\ \lambda - 3 & \lambda + 4 & 3\lambda \end{array} \right| \]

we will evaluate the determinant when \( \lambda = 0 \).

### Step 1: Substitute \( \lambda = 0 \)

Substituting \( \lambda = 0 \) in the determinant, we get:

\[ \left| \begin{array}{ccc} 0^2 + 3(0) & 0 - 1 & 0 + 3 \\ 0 + 1 & -2(0) & 0 - 4 \\ 0 - 3 & 0 + 4 & 3(0) \end{array} \right| = \left| \begin{array}{ccc} 0 & -1 & 3 \\ 1 & 0 & -4 \\ -3 & 4 & 0 \end{array} \right| \]

### Step 2: Calculate the determinant

Now we need to calculate the determinant:

\[ \left| \begin{array}{ccc} 0 & -1 & 3 \\ 1 & 0 & -4 \\ -3 & 4 & 0 \end{array} \right| \]

We can expand this determinant along the first row:

\[ = 0 \cdot \left| \begin{array}{cc} 0 & -4 \\ 4 & 0 \end{array} \right| - (-1) \cdot \left| \begin{array}{cc} 1 & -4 \\ -3 & 0 \end{array} \right| + 3 \cdot \left| \begin{array}{cc} 1 & 0 \\ -3 & 4 \end{array} \right| \]

### Step 3: Calculate the 2x2 determinants

1. For the second term:

\[ \left| \begin{array}{cc} 1 & -4 \\ -3 & 0 \end{array} \right| = (1)(0) - (-4)(-3) = 0 - 12 = -12 \]

2. For the third term:

\[ \left| \begin{array}{cc} 1 & 0 \\ -3 & 4 \end{array} \right| = (1)(4) - (0)(-3) = 4 - 0 = 4 \]

### Step 4: Substitute back into the determinant expansion

Now substituting back into the determinant expansion:

\[ = 0 + 12 + 3 \cdot 4 = 12 + 12 = 24 \]

### Step 5: Set the determinant equal to \( t \)

Since we have:

\[ t = 24 \]

Thus, the value of \( t \) is:

\[ \boxed{24} \]