Among the statements: = p x + q , then p 2 = 196 q 2 . [2026]

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Solution

Q.
Among the statements:

$I: If$ $| \begin{matrix} 1 & \cos α & \cos β \\ \cos α & 1 & \cos γ \\ \cos β & \cos γ & 1 \end{matrix} |$ $=$ $| \begin{matrix} 0 & \cos α & \cos β \\ \cos α & 0 & \cos γ \\ \cos β & \cos γ & 0 \end{matrix} |$ $, then \cos^{2} α + \cos^{2} β + \cos^{2} γ = \frac{3}{2},$ and

$II: If$ $| \begin{matrix} x^{2} + x & x + 1 & x - 2 \\ 2 x^{2} + 3 x - 1 & 3 x & 3 x - 3 \\ x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{matrix} |$ $= p x + q, then p^{2} = 196 q^{2} .$ [2026]

1 only I is true

2 both are false

3 only II is true

4 both are true

Ans.
(2)

$Let \cos α = x$

$\cos β = y$

$\cos γ = z$

$| \begin{matrix} 0 & x & y \\ x & 0 & z \\ y & z & 0 \end{matrix} |$ $=$ $| \begin{matrix} 1 & x & y \\ x & 1 & z \\ y & z & 1 \end{matrix} |$

Expanding both sides, we get

$x^{2} + y^{2} + z^{2} = 1$

$i.e. \cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$

Statement 1 is false.

Now,

$| \begin{matrix} x^{2} + x & 1 + x & x - 2 \\ 2 x^{2} + 3 x - 1 & 3 x & 3 x - 3 \\ x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{matrix} |$ $= p x + q$

Put $x = 0$ both sides

$q =$ $| \begin{matrix} 0 & 1 & - 2 \\ - 1 & 0 & - 3 \\ 3 & - 1 & - 1 \end{matrix} |$

$\Rightarrow q = - 12$

Now put $x = 1$ both sides

$p + q =$ $| \begin{matrix} 2 & 2 & - 1 \\ 4 & 3 & 3 \\ 6 & 1 & 1 \end{matrix} |$ $= 42$

$\Rightarrow p = 54$

Now,

$\frac{p^{2}}{q^{2}} = {(\frac{54}{- 12})}^{2} + 196$

$\Rightarrow p^{2} \neq 196 q^{2}$

Statement (2) is false.

Correct option (2).

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