Why does when three planes intersect in a line, $\Delta_x=\Delta_y=\Delta_z=0$?

Considering three equations $$\begin{cases}a_1x+b_1y+c_1z=d_1\\a_2x+b_2y+c_2z=d_2\\a_3x+b_3y+c_3z=d_3\end{cases}$$ let $$\Delta=\begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix} \Delta_x=\begin{vmatrix}d_1&b_1&c_1\\d_2&b_2&c_2\\d_3&b_3&c_3\end{vmatrix} \Delta_y=\begin{vmatrix}a_1&d_1&c_1\\a_2&d_2&c_2\\a_3&d_3&c_3\end{vmatrix} \Delta_z=\begin{vmatrix}a_1&b_1&d_1\\a_2&b_2&d_2\\a_3&b_3&d_3\end{vmatrix}$$ if the three planes intersect in a line, then $$\Delta=\Delta_x=\Delta_y=\Delta_z=0$$

I know the reason why $$\Delta=0$$, but I don't understand why$$\Delta_x=\Delta_y=\Delta_z=0$$

My thought is that if one of them doesn't equal to $$0$$, then the equations will have no solution, so they must equal to $$0$$

But can we explain that by using properties of determinant?

Multiply the given three eqns. by $$\vec i, \vec j, \vec k$$ we get $$\vec A x+ \vec B y +\vec C z= \vec D,~~ \vec A=a_1\vec i+ a_2 \vec j+ a_3 \vec k, etc.$$ Take cross multiplication of this eqn. by $$(\vec B \times \vec C)$$ from left to get $$[\vec A, \vec B, \vec C]x=[\vec D, \vec B,\vec C]\implies \Delta x= \Delta_1$$ Hwre $$[\vec P, \vec Q, \vec R]$$ is the scalar (box) product of three vectors which is alwaysa determinant. Similarly, we get $$\Delta y= \Delta_2$$ and $$\Delta z=\Delta_3$$ so $$\Delta=\Delta_1=\Delta_2=\Delta_3=0$$ lead to a consistent result $$0=0$$ three times Hence the three equation are consistent and they will have at least one solution. There will be only one (unique) solution if $$\Delta \ne 0$$, other wise many solutions. If $$\Delta=0$$ but any one of $$\Delta_1, \Delta_2, \Delta_3$$ is non-xero, it will lead to inconsistency so no solution.
Consider the matrix $$A = \begin{bmatrix} \vec{a} & \vec{b} & \vec{c} \\ \end{bmatrix}$$
$$K = \begin{bmatrix} \textbf{A} & \vec{d} \\ \end{bmatrix}$$ This Matrix has $$\text{dim}$$ $$\mathbb{N}(A) = 1$$, therefore it's column space has $$\text{dim}\ \text{Col}(A) = 2$$. I.e two rows are linearly independent. If $$\Delta_x$$ was not $$0$$ then it would follow that $$\vec{d}$$ is independent of $$\vec{b}$$ and $$\vec{c}$$ and that $$b$$ and $$c$$ are the Linearly independent of each other. This also implies that $$a$$ must be written as a combination of $$b$$ and $$c$$. since $$dim\ \text{col} (A) = 2$$. But since $$\textbf{d}$$ belongs to the column space of $$A = [ \vec{a} \ \ \vec{b} \ \ \vec{c} ]$$ it follows that the $$\vec{d}$$ is also expressible as a combination of $$\vec{b}$$ and $$\vec{c}$$ , which is a contradiction to $$\Delta_x = 0$$