Let $\alpha, \beta$ and $\gamma$ be real numbers such that the system of linear equations $$…

$x+2y+3z=\alpha$

$4x+5y+6z=\beta$

$7x+8y+9z=\gamma -1$

The system of linear equations is consistent it means it has unique solution or infinite solutions

Here, $\Delta =\left|\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right|=0$

Hence, there are infinitely many solutions.

If equations have infinitely many solutions, then the equations are linearly connected, i.e., $L_1+\lambda L_2=L_3$

$\Rightarrow \left(x+2y+3z-\alpha \right)+\lambda (4x+5y+6z-\beta )=7x+8y+9z-\gamma +1$

$\frac{1+4\lambda }{7}=\frac{2+5\lambda }{8}=\frac{3+6\lambda }{9}=\frac{\alpha +\lambda \beta }{\gamma -1}$

$\frac{1+4\lambda }{7}=\frac{2+5\lambda }{8}\Rightarrow \lambda =-2$

Also,$\frac{1+4\lambda }{7}=\frac{\alpha +\lambda B}{\gamma -1}$

$\Rightarrow -1=\frac{\alpha -2\beta }{r-1}$

$\Rightarrow \alpha -2\beta +\gamma =1$

Now, $P$ is the plane containing the points $\left(\alpha ,\beta ,\gamma \right)$

So, the equation of the plane is $x-2 y+z=1$ (replacing $\alpha , \beta , \gamma$ by $x, y, z$)

We know, the distance of a point $\left(x_1, y_1, z_1\right)$ from plane $ax+by+cz+d=0$ is $\left|\frac{a{x}_1+b{y}_1+c{z}_1+d}{\sqrt{a^2+b^2+c^2}}\right|$

So, $D={\left(\frac{0\times 1-2\times 1+0\times 1-1}{\sqrt{1^2+(-2{)}^2+1^2}}\right)}^2=\frac{9}{6}=1.50$