Explanation

(c)
We have,
$ax^{4} + bx^{3} + cx^{2} + 50x + d = \left| \begin{matrix}
x^{3} - 14x^{2} & - x & 3x + \lambda \\
4x + 1 & 3x & x - 4 \\
- 3 & 4 & 0
\end{matrix} \right|$
On differentiating with respect to $x$, we get
$4ax^{3} + 3bx^{2} + 2cx + 50 = \left| \begin{matrix}
3x^{2} - 28x & - 1 & 3 \\
4x + 1 & 3x & x - 4 \\
- 3 & 4 & 0
\end{matrix} \right| + \left| \begin{matrix}
x^{3} - 14x^{2} & - x & 3x + \lambda \\
4 & 3 & 1 \\
- 3 & 4 & 0
\end{matrix} \right|$
Now, put $x = 0$, we get
$50 = \left| \begin{matrix}
0 & - 1 & 3 \\
1 & 0 & - 4 \\
- 3 & 4 & 0
\end{matrix} \right| + \left| \begin{matrix}
0 & 0 & \lambda \\
4 & 3 & 1 \\
- 3 & 4 & 0
\end{matrix} \right|$
$\Rightarrow 50 = 25\lambda$
$\Rightarrow \ \lambda = 2$