\(\quad 12-1012:(\pi / 9) \int_{0}(\tan x+\tan 2 x+\tan 3 x+\tan x \cdot \tan 2 x \cdot \tan 3 x) d x\) is equal to....... (a) \((1 / 3) \log 2\) (b) \(\log ^{3} \sqrt{4}\) (c) \(3 \log 2\) (d) \(4 \log \sqrt{3}\)

Sum and product of integrals can be expressed as multiple integrals with same limits: \[12 - 1012\left(\frac{\pi}{9}\right)\int_{0}^{\pi/9}(\tan x+\tan 2x+\tan 3x+\tan x \cdot \tan 2x \cdot \tan 3x) dx\] \[= 12 - 1012\left(\frac{\pi}{9}\right)\left[\int_{0}^{\pi/9}\tan xdx+\int_{0}^{\pi/9}\tan2xdx+\int_{0}^{\pi/9}\tan3xdx+\int_{0}^{\pi/9}\tan x \cdot \tan 2x \cdot \tan 3x dx\right]\]

Using antiderivative formulas, we can evaluate each integral: \[\int_{0}^{\pi/9}\tan x dx = \log|\sec x| \Big|_0^{\pi/9}\] \[\int_{0}^{\pi/9}\tan2x dx = \frac{1}{2}\log|\sec 2x| \Big|_0^{\pi/9}\] \[\int_{0}^{\pi/9}\tan3x dx = \frac{1}{3}\log|\sec 3x| \Big|_0^{\pi/9}\] For the product term, use the triple-angle formula for the tangent: \[\tan 3x = 3\tan x -\tan^3 x\] Rearrange it as: \[\tan x\tan 2x\tan 3x = (\tan^3 x)\left(\frac{\tan 3x}{\tan x}\right) = \tan^3x (3-\tan^3x)\] Now evaluate this integral: \[\int_{0}^{\pi/9}\tan x \cdot \tan 2x \cdot \tan 3x dx = \int_{0}^{\pi/9}(\tan^3x(3-\tan^3x)) dx\] This last integral is complicated to evaluate directly, but notice that at \(x = \pi/9\), both \(\tan x\) and \(\tan 2x\) are equal to 1, so their product with any other term will still be 0. Therefore, the integral is zero: \[\int_{0}^{\pi/9}\tan x \cdot \tan 2x \cdot \tan 3x dx = 0\]

Substitute the evaluated integrals: \[12 - 1012\left(\frac{\pi}{9}\right)\left[ \log|\sec x| \Big|_0^{\pi/9} + \frac{1}{2}\log|\sec 2x| \Big|_0^{\pi/9} + \frac{1}{3}\log|\sec 3x| \Big|_0^{\pi/9} + 0 \right]\] Now evaluate the difference at the limits: \[12 - 1012\left(\frac{\pi}{9}\right)\left[ (\log|\sec(\pi/9)| - \log|\sec0|) + \frac{1}{2}(\log|\sec(2\pi/9)| - \log|\sec0|) + \frac{1}{3}(\log|\sec(\pi/3)| - \log|\sec0|) \right]\] Evaluate the secant values and simplify the logarithms: \(12 - 1012\left(\frac{\pi}{9}\right)\left[ \log2 + \frac{1}{2}\log2 + \frac{1}{3}(\log2 + \log3 - \log1) \right]\) Combine all logarithm terms: \[12 - 1012\left(\frac{\pi}{9}\right)\left[\log2\left(1 + \frac{1}{2}\right) + \frac{1}{3}\log6\right]\] \[12 - 1128\log2 - 376\log3\] Now check the options: (a) \(\frac{1}{3}\log2 \approx 0.23\) (b) \(\log ^{3} \sqrt{4} \approx 0.602\) (c) \(3\log2 \approx 2.08\) (d) \(4\log\sqrt{3} \approx 2.191\) Our result, \(12 - 1128\log2 - 376\log3\), does not match any of the given options. Therefore, the problem might be ill-posed or we used an incorrect technique. Either way, the answer cannot be determined from the given information.