Consider the planes $P_1:2x+y+z+4=0, P_2:y–z+4=0$and $P_3 : 3x+2y+z+8=0.$ Let $L_1, L_2, L_3$ be the lines of intersection of the planes $P_2$ and $P_3, P_3$ and $P_1$, and $P_1$ and $P_2$ respectively. Then,

Observe that the lines $L_1, L_2 and L_3$ are parallel to the vector $\widehat{\text{i}}-\widehat{\text{j}}-\widehat{\text{k}}$

Also, $\Delta =0={\Delta }_1 \& b_1{c}_2 \ne b_1{c}_1$ Hence the three planes intersect in a line.
$P_1=2x+y+z+4=0$

$P_2$ = 0x + y – z + 4 = 0
$P_3$ = 3x + 2y + z + 8 = 0
$P_2$ and $P_3$ gives line L1
vector parallel to line $L_1=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ 0& 1& -1\\ 3& 2& 1\end{array}\right|$
$$\begin{gathered} =3\widehat{i}-3\widehat{j}-3\widehat{k} \\ =3[\widehat{i}-\widehat{j}-\widehat{k}] \end{gathered}$$

Similarly
vector parallel to $L_3$, $P_1$ and $P_2=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ 2& 1& 1\\ 0& 1& -1\end{array}\right|$
$$\begin{gathered} =(-2)\widehat{i}-(-2)\widehat{j}+2\widehat{k} \\ =-2\widehat{i}+2\widehat{j}+2\widehat{k} \end{gathered}$$

$=-2(\widehat{i}-\widehat{j}-\widehat{k})$

We can see all the lines are parallel to vector $(\widehat{i}-\widehat{j}-\widehat{k})$
Also 2x + y + z = – 4
0x + y – z = – 4
3x + y + z = – 8
$$\begin{gathered} \Delta =\left|\begin{array}{ccc}2& 1& 1\\ 0& 1& -1\\ 3& 2& 1\end{array}\right| \\ \Rightarrow 2(1+2)-1(0+3)+1(0-3) \\ \Rightarrow 2\left(3\right)-3-3=0{\Delta }_1=\left|\begin{array}{ccc}-4& 1& 1\\ -4& 1& 1\\ -8& 2& 1\end{array}\right|=0 \\ {\Delta }_2=\left|\begin{array}{ccc}2& -4& 1\\ 0& -4& -1\\ 3& -8& 1\end{array}\right|=0 \\ =-4\left|\begin{array}{ccc}2& 1& 1\\ 0& 1& -1\\ 3& 2& 1\end{array}\right|=0 \\ {\Delta }_3=0 \end{gathered}$$

So all planes intersection in line L.