The number of real solutions of the equation $$3\left( {{x^2} + {1 \over {{x^2}}}} \right) - 2\left( {x + {1 \over x}} \right) + 5 = 0$$, is

The number of integers, greater than 7000 that can be formed, using the digits 3, 5, 6, 7, 8 without repetition is :

Let $$\overrightarrow \alpha = 4\widehat i + 3\widehat j + 5\widehat k$$ and $$\overrightarrow \beta = \widehat i + 2\widehat j - 4\widehat k$$. Let $${\overrightarrow \beta _1}$$ be parallel to $$\overrightarrow \alpha $$ and $${\overrightarrow \beta _2}$$ be perpendicular to $$\overrightarrow \alpha $$. If $$\overrightarrow \beta = {\overrightarrow \beta _1} + {\overrightarrow \beta _2}$$, then the value of $$5{\overrightarrow \beta _2}\,.\left( {\widehat i + \widehat j + \widehat k} \right)$$ is :

If the system of equations

$$x+2y+3z=3$$

$$4x+3y-4z=4$$

$$8x+4y-\lambda z=9+\mu$$

has infinitely many solutions, then the ordered pair ($$\lambda,\mu$$) is equal to :