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Consider the lines

\(L_{1}\) : \(\frac{x+1}{3}\) = \(\frac{y+2}{1}\) = \(\frac{z+1}{2}\)


\(L_{2}\) : \(\frac{x-2}{1}\) = \(\frac{y+2}{2}\) = \(\frac{z-3}{3}\)

(Following Equations Represents Equation of Line In 3 Dimensional Cartesian Coordinate System)

If the unit vector perpendicular to both \(L_{1}\) and \(L_{2}\) can be written as

\(\frac{a\hat{i}-b\hat{j}+c\hat{k}}{d\sqrt{e}}\) ; where \(\hat{i}, \hat{j}, \hat{k}\) are unit vectors along x-axis, y-axis, z-axis respectively and a,b,c,d,e \(\in\) R

*Now if \(f(x)\)a function which is differentiable and *

\(\int_0^{t^{2}}xf(x)dx\) = \(\frac{2}{5}t^{5}\)

then \(f(\frac{e-a}{cd}\)) = \(\frac{\alpha}{\beta}\)

where \(\beta > \alpha\)

Then value of \( 1 + \omega^{\alpha + \beta} + \omega^{2b}\) =

(Where \(\omega\) is imaginary cube root of unity)


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