MCQ
If $ \displaystyle \begin{vmatrix} \text{x} &\text{amp;}\text{ y} \\ 1 &\text{amp; } 6 \end{vmatrix}= \displaystyle \begin{vmatrix} 1&\text{amp; }8 \\ 1 &\text{amp; } 6 \end{vmatrix}$ then $\text{x}+2\text{y}=$
- A$9$
- ✓$17$
- C$10$
- D$7$
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$(a-c) x^2+(b-a) x+(c-b)=0$ where $a, b, c$ are distinct real numbers such that the matrix
$\left[\begin{array}{ccc}\alpha^2 & \alpha & 1 \\1 & 1 & 1 \\a & b & c\end{array}\right]$
is singular. Then the value of
$\frac{(a-c)^2}{(b-a)(c-b)}+\frac{(b-a)^2}{(a-c)(c-b)}+\frac{(c-b)^2}{(a-c)(b-a)}$