@srestha ma'am see this

Given that $A=\begin{bmatrix} 0 &4 &2 &1 \\ 3 &-1 &0 &2 \\ 5 &2 &x &4 \\ 6 &1 &-1 &0 \end{bmatrix}$ and $|A|=245$

find $x=?$

Now $|A|=\begin{vmatrix} 0 &4 &2 &1 \\ 3 &-1 &0 &2 \\ 5 &2 &x &4 \\ 6 &1 &-1 &0 \end{vmatrix}$

expand the row wise$,$

$|A|=0.\begin{vmatrix} -1 &0 &2 \\ 2 &x &4 \\ 1 &-1 &0 \end{vmatrix}$$-4.\begin{vmatrix} 3&0 &2 \\ 5 &x &4 \\ 6 &-1 &0 \end{vmatrix}$$+2.\begin{vmatrix} 3&-1 &2 \\ 5 &2 &4 \\ 6 &1 &0 \end{vmatrix}$$-1.\begin{vmatrix} 3&-1 &0 \\ 5 &2 &x \\ 6 &1 &-1 \end{vmatrix}$

$|A|=0-4.\begin{vmatrix} 3&0 &2 \\ 5 &x &4 \\ 6 &-1 &0 \end{vmatrix}$$+2.\begin{vmatrix} 3&-1 &2 \\ 5 &2 &4 \\ 6 &1 &0 \end{vmatrix}$$-1.\begin{vmatrix} 3&-1 &0 \\ 5 &2 &x \\ 6 &1 &-1 \end{vmatrix}$

$|A|=-4.\begin{vmatrix} 3&0 &2 \\ 5 &x &4 \\ 6 &-1 &0 \end{vmatrix}$$+2.\begin{vmatrix} 3&-1 &2 \\ 5 &2 &4 \\ 6 &1 &0 \end{vmatrix}$$-1.\begin{vmatrix} 3&-1 &0 \\ 5 &2 &x \\ 6 &1 &-1 \end{vmatrix}$

Now,$|A|=-4|A_{1}|+2|A_{2}|-1|A_{3}|$

$|A|=-4|A_{1}|+2|A_{2}|-|A_{3}|$--------->$(1)$

Where $|A_{1}|=\begin{vmatrix} 3&0 &2 \\ 5 &x &4 \\ 6 &-1 &0 \end{vmatrix}=3(0+4)-0(0-24)+2(-5-6x)=2-12x$

$|A_{2}|=\begin{vmatrix} 3&-1 &2 \\ 5 &2 &4 \\ 6 &1 &0 \end{vmatrix}=3(0-4)+1(0-24)+2(5-12)=-12-24-14=-50$

$|A_{3}|=\begin{vmatrix} 3&-1 &0 \\ 5 &2 &x \\ 6 &1 &-1 \end{vmatrix}=3(-2-x)+1(-5-6x)=-6-3x-5-6x=-9x-11$

Put these values in equation $(1),$

Now$,|A|=-4(2-12x)+2(-50)-(-9x-11)$

Given that $|A|=245$

$\Rightarrow -8+48x-100+9x+11=245$

$\Rightarrow 57x-100+3=245$

$\Rightarrow 57x-97=245$

$\Rightarrow 57x=245+97$

$\Rightarrow 57x=342$

$\Rightarrow x=\frac{342}{57}$

$\Rightarrow x=6$

So$,x=6$ is the correct answer.

But this is a time-consuming method. Try, to do some operation in rows or columns and make some rows and columns $0$, then it will be easy.