The distance of the point (1, 3, – 7) from the plane passing through the point (1, –1, – 1), having normal perpendicular to both the lines

$${{x - 1} \over 1} = {{y + 2} \over { - 2}} = {{z - 4} \over 3}$$

and

$${{x - 2} \over 2} = {{y + 1} \over { - 1}} = {{z + 7} \over { - 1}}$$ is :

The eccentricity of an ellipse whose centre is at the origin is $${1 \over 2}$$. If one of its directrices is x = – 4, then the equation of the normal to it at $$\left( {1,{3 \over 2}} \right)$$ is :

The radius of a circle, having minimum area, which touches the curve y = 4 – x² and the lines, y = |x| is :

Let $$\overrightarrow a = 2\widehat i + \widehat j -2 \widehat k$$ and $$\overrightarrow b = \widehat i + \widehat j$$.

Let $$\overrightarrow c $$ be a vector such that $$\left| {\overrightarrow c - \overrightarrow a } \right| = 3$$,

$$\left| {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c } \right| = 3$$ and the angle between $$\overrightarrow c $$ and $\overrightarrow a \times \overrightarrow b$ is $$30^\circ $$.

Then $$\overrightarrow a .\overrightarrow c $$ is equal to :