The value of D is .........

The given system in the variables $$x,y,z$$ is

$$\begin{aligned} x + 2y + 3z &= \alpha \\[2pt] 4x + 5y + 6z &= \beta \\[2pt] 7x + 8y + 9z &= \gamma-1 \end{aligned}$$

Write the coefficient matrix $$A$$ and the right-hand side vector $$b$$:

$$A=\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix},\qquad b=\begin{bmatrix}\alpha\\ \beta\\ \gamma-1\end{bmatrix}$$

The system is consistent exactly when $$b$$ lies in the column space of $$A$$. Therefore $$\text{rank}(A)=\text{rank}\!\left([A\; b]\right).$$

First find the rank of $$A$$. Compute its determinant:

$$\det A =1(5\cdot9-6\cdot8)-2(4\cdot9-6\cdot7)+3(4\cdot8-5\cdot7) =-3+12-9 = 0$$

Thus $$\det A=0$$, so $$\text{rank}(A)\le 2$$. Indeed, the rows are linearly dependent, because

$$\mathbf{r}_1-2\mathbf{r}_2+\mathbf{r}_3=0,$$

where $$\mathbf{r}_1=(1,2,3),\; \mathbf{r}_2=(4,5,6),\; \mathbf{r}_3=(7,8,9).$$

Hence $$\text{rank}(A)=2$$. For consistency the same linear relation must hold for the entries of $$b$$:

$$1\cdot\alpha-2\cdot\beta+1\cdot(\gamma-1)=0$$

$$\Longrightarrow\; \alpha-2\beta+\gamma = 1\; .$$

All triples $$\left(\alpha,\beta,\gamma\right)$$ satisfying this equation lie on the plane

$$P:\; \alpha-2\beta+\gamma-1=0.$$

The problem asks for $$D$$, the square of the distance from the point $$(0,1,0)$$ to the plane $$P$$.

The distance from a point $$(\alpha_0,\beta_0,\gamma_0)$$ to a plane $$A\alpha+B\beta+C\gamma+D=0$$ is

$$d=\frac{\lvert A\alpha_0+B\beta_0+C\gamma_0+D\rvert} {\sqrt{A^{2}+B^{2}+C^{2}}}.$$

For plane $$P$$ we have $$A=1,\; B=-2,\; C=1,\; D=-1$$. Insert the coordinates of the point $$(0,1,0)$$:

$$d=\frac{\lvert 1\cdot0+(-2)\cdot1+1\cdot0-1\rvert} {\sqrt{1^{2}+(-2)^{2}+1^{2}}} =\frac{\lvert -2-1\rvert}{\sqrt{1+4+1}} =\frac{3}{\sqrt{6}} =\frac{\sqrt{6}}{2}.$$

Therefore

$$D=d^{2}=\left(\frac{\sqrt{6}}{2}\right)^{2}=\frac{6}{4}=\frac{3}{2}=1.5.$$

Hence the required value lies in the range $$1.45-1.55$$.