$\left|\begin{matrix} 0 & 4 & 4\\3 & 2 & 4\\1 & 8 & 11\end{matrix}\right|+\left|\begin{matrix} 0 & 4 & 4\\-3 & 5 & 0\\1 & 4 & 7\end{matrix} \right|+4\left|\begin{matrix} 1 & 1 & 1\\1 & 7 & 4\\0 & 4 & 7\end{matrix}\right|$

Apply $R_{3}\rightarrow R_{3}-R_{1}$ on first determinant and multiply scalar $4$ to first row of third determinant.

$\left|\begin{matrix} 0 & 4 & 4\\3 & 2 & 4\\1 & 4 & 7\end{matrix}\right|+\left|\begin{matrix} 0 & 4 & 4\\-3 & 5 & 0\\1 & 4 & 7\end{matrix} \right|+\left|\begin{matrix} 4 & 4 & 4\\1 & 7 & 4\\0 & 4 & 7\end{matrix}\right|$

Note that Row 1 and Row 3 in first and second matrix are same, so we can add second row to add determinants.

$\left|\begin{matrix} 0 & 4 & 4\\0 & 7 & 4\\1 & 4 & 7\end{matrix}\right|+\left|\begin{matrix} 4 & 4 & 4\\1 & 7 & 4\\0 & 4 & 7\end{matrix}\right|$

Now, second and third columns are same, so we can add column one!

$\left|\begin{matrix} 4 & 4 & 4\\1 & 7 & 4\\1 & 4 & 7\end{matrix}\right|$

Solving this determinant, $4(49-16)-4(7-4)+4(4-7)=108$

Edit: If we have to do it by triangular matrix and not expanding,

$4\begin{vmatrix} 1 & 1 & 1 \\ 1 & 7 & 4 \\ 1 & 4 & 7 \end{vmatrix}$

${ C }_{ 2 }=C_{ 1 }\times (-1)+{ C }_{ 2 }\\ { C }_{ 3 }=C_{ 1 }\times (-1)+{ C }_{ 3 }$

$4\begin{vmatrix} 1 & 0 & 0 \\ 1 & 6 & 3 \\ 1 & 3 & 6 \end{vmatrix}$

${ C }_{ 3 }=C_{ 2 }\times (\frac { -1 }{ 2 } )+{ C }_{ 3 }$

$4\begin{vmatrix} 1 & 0 & 0 \\ 1 & 6 & 0 \\ 1 & 3 & 4.5 \end{vmatrix}$

$4\times [1\times 6\times 4.5]=\boxed { 108 }$

I find using row operations to reduce the final calculation is very effective. There are only two rules to it.

1. The value of determinant does not change if we replace a $row_i$ with $row_i - C \dot{}row_j$ .
2. When two rows are swapped the resultant determinant is multiplied by $-1$ .

For $3\times 3$ matrix, the calculations are much reduced if we create two zeros in a column. Let us look at the example provided by this problem.

$X = \begin{vmatrix} 0&4&4 \\ 3&2&4 \\ 1&8&11 \end{vmatrix} + \begin{vmatrix} 0&4&4 \\ -3&5&0 \\ 1&4&7 \end{vmatrix} + 4 \begin{vmatrix} 1&1&1\\ 1&7&4\\ 0&4&7 \end{vmatrix}$

$\quad = \begin{vmatrix} 0&4&4 \\ 0&-22&-29 \\ 1&8&11 \end{vmatrix} + \begin{vmatrix} 0&4&4 \\ 0&17&21 \\ 1&4&7 \end{vmatrix} + 4 \begin{vmatrix} 1&1&1\\ 0&6&3\\ 0&4&7 \end{vmatrix}$

$\quad = (4)(-29)(1)-(4)(-22)(1)+(4)(21)(1)$

$\quad \quad -(4)(17)(1)+4[(1)(6)(7)-(1)(3)(4)]$

$\quad = -116- (-88)+84-68+4(42-12)$

$\quad =-28+16+4(30) = \boxed{108}$

$X=\left| \begin{matrix} 0 & 4 & 4 \\ 3 & 2 & 4 \\ 1 & 8 & 11 \end{matrix} \right| +\left| \begin{matrix} 0 & 4 & 4 \\ -3 & 5 & 0 \\ 1 & 4 & 7 \end{matrix} \right| +4\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 7 & 4 \\ 0 & 4 & 7 \end{matrix} \right|$

$1^{st} determinant - R_{1} ~changes~ to R_{2}$

$X= - \left| \begin{matrix} 3 & 2 & 4 \\ 0 & 4 & 4 \\ 1 & 8 & 11 \end{matrix} \right|$

$R_{3} \to R_{3} - \dfrac{1}{3}R_{1}$

$X= - \left| \begin{matrix} 3 & 2 & 4 \\ 0 & 4 & 4 \\ 0 & \dfrac{22}{3} & \dfrac{29}{3} \end{matrix} \right|$

Thus triangular matrix , therefore determinant = multiplication of the diagonals =

$- (3\times4\times7/3) = -28$

(I am not showing this , try yourself)

Similarly doing for second(as done for first one) we get elements in the main diagonal , thus determinant =

$- (-3\times 4 \times 4/3 = 16$

For third ,

$\left|\begin{matrix} 1 & 1 & 1 \\ 1 & 7 & 4 \\ 0 & 4 & 7 \end{matrix} \right|$

$R_{2} \to R_{2} - R_{1}$

$\left|\begin{matrix} 1 & 1 & 1 \\ 0 & 6 & 3 \\ 0 & 4 & 7 \end{matrix} \right|$

$R_{3} \to R_{3} - \dfrac{4}{6} \times R_{2}$

$\left|\begin{matrix} 1 & 1 & 1 \\ 1 & 7 & 4 \\ 0 & 0 & 5 \end{matrix} \right|$

once again triangular matrix therefore determinant = $1\times6\times5 = 30$

$X = -28 + 16 + 4(30) = 108$

basket method is one of the easiest solution to this kind of prblem