If $I_{1}, I_{2}, I_{3}$ are the intercepts of $x^{2} + y^{2} - 14 x - 10 y + 24 = 0$ on $x -$ axis, $y -$ axis and $y = x$ respectively, then

I_{2} > I_{3} > I_{1}

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I_{3} > I_{1} > I_{2}

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I_{1} > I_{2} > I_{3}

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I_{3} > I_{2} > I_{1}

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Solution

The correct option is B $I_{3} > I_{1} > I_{2}$
$S : x^{2} + y^{2} - 14 x - 10 y + 24 = 0$
To find $x -$ intercept, substitute $(y = 0)$
$x^{2} - 14 x + 24 = 0$
$I_{1} = (x_{1} - x_{2}) = \sqrt{(14)^{2} - 4 \times 24}$
$= \sqrt{196 - 96}$
$I_{1} = 10$
To find $y -$ intercept, substitute $(x = 0)$
$y^{2} - 10 y + 24 = 0$
$I_{2} = | y_{1} - y_{2} | = \sqrt{(10)^{2} - 4 \times 24}$
$I_{2} = 2$
Intercept of $y = x$
$2 x^{2} - 24 x + 24 = 0$
$x^{2} - 12 x + 12 = 0$
$(x_{1} - x_{2}) = \sqrt{(12)^{2} - 4 \times 12}$
$= \sqrt{96}$
$∴$ Intercept of $y = x$ is $I_{3} = \sqrt{2} | x_{1} - x_{2} | = \sqrt{2} | y_{1} - y_{2} |$
$= \sqrt{2} \sqrt{96}$
$I_{3} = \sqrt{192}$
$∴ I_{3} > I_{1} > I_{2}$

Hence, option B.

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Now what should be the correct relation between $I_{1}$ , $I_{2}$ and $I_{3}$ .

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