The combined equation of the two lines $$ax+by+c=0$$ and $$a'x+b'y+c'=0$$ can be written as

$$(ax+by+c)(a'x+b'y+c')=0$$.

The equation of the angle bisectors of the lines represented by the equation $$2x^2+xy-3y^2=0$$ is :

In a binomial distribution $$B(n,p)$$, the sum and the product of the mean and the variance are 5 and 6 respectively, then $$6(n+p-q)$$ is equal to :

Let $$S = \left\{ {x:x \in \mathbb{R}\,\mathrm{and}\,{{(\sqrt 3 + \sqrt 2 )}^{{x^2} - 4}} + {{(\sqrt 3 - \sqrt 2 )}^{{x^2} - 4}} = 10} \right\}$$. Then $$n(S)$$ is equal to

If the center and radius of the circle $$\left| {{{z - 2} \over {z - 3}}} \right| = 2$$ are respectively $$(\alpha,\beta)$$ and $$\gamma$$, then $$3(\alpha+\beta+\gamma)$$ is equal to :