If the matrix $\left[ \begin{matrix}
   1 & 3 & \lambda +2 \\
   2 & 4 & 8 \\
   3 & 5 & 10 \\
\end{matrix} \right]$ is singular, then $\lambda $ =
A . -2
B . 4
C . 2
D . -4

Hint: We are given a matrix which is of order $3\times 3$and given that the matrix is a singular matrix. That means the determinant of the matrix is equal to zero. Thus, first we find the determinant of the given matrix and then we put it equal to zero and after simplifying it, we are able to get the value and choose the correct option.

Complete Step- by- step Solution:
Given matrix is $\left[ \begin{matrix}
   1 & 3 & \lambda +2 \\
   2 & 4 & 8 \\
   3 & 5 & 10 \\
\end{matrix} \right]$
As the matrix is of the order $3\times 3$, so it contains 3 rows and 3 columns.
Now we find the determinant of the given square matrix and then equate it to zero.
We find the determinant by using the first row.
The determinant is
$[1[4\times 10-5\times 8]-3[2\times 10-3\times 8]+[\lambda +2][2\times 5-3\times 4]]$
Simplifying the above equation, we get
$[1[40-40]-3[20-24]+[\lambda +2][10-12]]$
Evaluating further, we get
$[12-2\lambda -4]$
Now we equate the above equation equal to zero [ as it is given to be singular matrix]
We get $-2\lambda +8=0$
Hence $\lambda =4$
Hence the value of $\lambda =4$

Thus, Option (B) is the correct answer.

Note: Students must remember that the matrix should have same number of rows and columns to find the determinant of the matrix. This means we can find the determinant of the square matrix. We should have the practice of finding the determinant to solve the question in lesser time.