Mathematics · 2020
JEE Main · 5 September 2020, Shift 2 · Q53
If a +x= b +y= c +z+1, where a, b, c, x, y, z are non-zero distinct real numbers, then |x, a +y, x+ a; y, b +y, y+ b; z, c +y, z+ c| is equal to:
If $\displaystyle \mathrm{a}+x=\mathrm{b}+y=\mathrm{c}+z+1$, where $\displaystyle \mathrm{a}, \mathrm{b}, \mathrm{c}, x$, $\displaystyle y, z$ are non-zero distinct real numbers, then $\displaystyle \left|\begin{array}{lll}x & \mathrm{a}+y & x+\mathrm{a} \\ y & \mathrm{~b}+y & y+\mathrm{b} \\ z & \mathrm{c}+y & z+\mathrm{c}\end{array}\right|$ is equal to :
Official answer
(2)
$\displaystyle y(\mathrm{a}-\mathrm{b})$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.