2021-M2-08

$\begin{array}{rcl}
\begin{vmatrix} 1 & d-1 & d+3 \\ 2 & d+2 & -1 \\ 3 & d+4 & 5 \end{vmatrix} & = & 0 \\
(1) \begin{vmatrix} d+2 & -1 \\ d+4 & 5 \end{vmatrix}-(2) \begin{vmatrix} d-1 & d+3 \\ d+4 & 5 \end{vmatrix}+(3)\begin{vmatrix} d-1 & d+3 \\ d+2 & -1 \end{vmatrix} & = & 0 \\
5(d+2)+(d+4)-2[5(d-1)-(d+3)(d+4)] +3[-(d-1)-(d+3)(d+2)] & = & 0 \\
5d+10+d+4-10d+10+2d^2+14d+24-3d+3-3d^2-15d-18 & = & 0 \\
-d^2-8d+33 & = & 0 \\
(d-3)(d+11) & = & 0
\end{array}$

$\therefore d=3$ or $d=-11$.

For $d=3$, we have

$\begin{array}{l}
\left(\begin{array}{ccc|c} 1 & 2 & 6 & 1 \\ 2 & 5 & -1 & 1 \\ 3 & 7 & 5 & 2 \end{array}\right) \\
\xrightarrow[R_3-3R_1\to R_3]{R_2-2R_1\to R_2} \left(\begin{array}{ccc|c} 1 & 2 & 6 & 1 \\ 0 & 1 & -13 & -1 \\ 0 & 1 & -13 & -1 \end{array}\right) \\
\xrightarrow{R_3-R_2 \to R_3} \left(\begin{array}{ccc|c} 1 & 2 & 6 & 1 \\ 0 & 1 & -13 & -1 \\ 0 & 0 & 0 & 0 \end{array}\right)
\end{array}$

Let $z=t$, where $t\in\mathbb{R}$.

$\begin{array}{rcl}
y-13t & = & -1 \\
y & = & 13t-1
\end{array}$

$\begin{array}{rcl}
x+2(13t-1)+6t & = & 1 \\
x & = & 3-32t
\end{array}$

Therefore, the solution of $(E)$ is $\{ (3-32t, 13t-1, t) | t \in \mathbb{R} \}$.

For $d=-11$, we have

$\begin{array}{l}
\left(\begin{array}{ccc|c} 1 & -12 & -8 & 15 \\ 2 & -9 & -1 & -27 \\ 3 & -7 & 5 & 2 \end{array}\right) \\
\xrightarrow[R_3-3R_1\to R_3]{R_2-2R_1\to R_2} \left(\begin{array}{ccc|c} 1 & -12 & -8 & 15 \\ 0 & 15 & 15 & -57 \\ 0 & 29 & 29 & -43 \end{array}\right) \\
\xrightarrow[R_3-R_2\times \frac{29}{15}\to R_3]{R_2\div 15 \to R_2} \left(\begin{array}{ccc|c} 1 & -12 & -8 & 15 \\ 0 & 1 & 1 & \dfrac{-57}{15} \\ 0 & 0 & 0 & \dfrac{336}{5} \end{array}\right)
\end{array}$

Therefore, $(E)$ has no solution if $d=-11$. Hence, $d \neq -11$.

$\begin{array}{rcl}
(3-32t)(13t-1)+2(3-32t)(t) & = & 3 \\
39t-3-416t^2+32t+6t-64t^2-3 & = & 0 \\
480t^2-77t+6 & = & 0 \\
t & = & \dfrac{-(-77)\pm\sqrt{(-77)^2-4(480)(6)}}{2(480)} \\
t & = & \dfrac{77\pm\sqrt{-5591}}{960} \text{ , which is not a real number.}
\end{array}$

Hence, $(E)$ does not have a real solution $(x,y,z)$ satisfying $xy+2xz=3$.

Therefore, the claim is not correct.