Given, ${\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right]}_{1\times 3}{\left[\begin{array}{ccc}1& 3& 2\\ 2& 5& 1\\ 15& 3& 2\end{array}\right]}_{3\times 3}{\left[\begin{array}{l}1\\ 2\\ \mathrm{x}\end{array}\right]}_{3\times 1}=\mathrm{O}$
$\Rightarrow \left[\begin{array}{l}1+2\mathrm{x}+15\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3+5\mathrm{x}+3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2+\mathrm{x}+2\end{array}\right]\left[\begin{array}{l}1\\ 2\\ \mathrm{x}\end{array}\right]=\mathrm{O}$

$\Rightarrow [16+2\mathrm{x}+153+5\mathrm{x}+32+\mathrm{x}+2]\left[\begin{array}{l}1\\ 2\\ \mathrm{x}\end{array}\right]=\mathrm{O}$
⇒ [(16 + 2x) ⋅ 1 + (6 + 5x) ⋅ 2 + (4 + x) ⋅ x] = 0

⇒ (16 + 2x) + (12 + 10x) + (4x + x²) = 0

⇒ x² + 16x + 28 = 0 ⇒ (x + 14) (x + 2) = 0

⇒ x + 14 = 0 or x + 2 = 0

Hence, x = – 14 or x = – 2.