Solution of the differential equation $$ydx + \left( {x + {x^2}y} \right)dy = 0$$ is

A particle acted on by constant forces $$4\widehat i + \widehat j - 3\widehat k$$ and $$3\widehat i + \widehat j - \widehat k$$ is displaced from the point $$\widehat i + 2\widehat j + 3\widehat k$$ to the point $$\,5\widehat i + 4\widehat j + \widehat k.$$ The total work done by the forces is :

If the straight lines
$$x=1+s,y=-3$$$$ - \lambda s,$$ $$z = 1 + \lambda s$$ and $$x = {t \over 2},y = 1 + t,z = 2 - t,$$ with parameters $$s$$ and $$t$$ respectively, are co-planar, then $$\lambda $$ equals :

Let $$\overrightarrow a ,\overrightarrow b $$ and $$\overrightarrow c $$ be three non-zero vectors such that no two of these are collinear. If the vector $$\overrightarrow a + 2\overrightarrow b $$ is collinear with $$\overrightarrow c $$ and $$\overrightarrow b + 3\overrightarrow c $$ is collinear with $$\overrightarrow a $$ ($$\lambda $$ being some non-zero scalar) then $$\overrightarrow a + 2\overrightarrow b + 6\overrightarrow c $$ equals to :