If $$\overrightarrow a ,\overrightarrow b $$ and $$\overrightarrow c $$ are unit vectors satisfying
$${\left| {\overrightarrow a - \overrightarrow b } \right|^2} + {\left| {\overrightarrow b - \overrightarrow c } \right|^2} + {\left| {\overrightarrow c - \overrightarrow a } \right|^2} = 9,$$ then $$\left| {2\overrightarrow a + 5\overrightarrow b + 5\overrightarrow c } \right|$$ is

Let $$\overrightarrow a = - \widehat i - \widehat k,\overrightarrow b = - \widehat i + \widehat j$$ and $$\overrightarrow c = \widehat i + 2\widehat j + 3\widehat k$$ be three given vectors. If $$\overrightarrow r $$ is a vector such that $$\overrightarrow r \times \overrightarrow b = \overrightarrow c \times \overrightarrow b $$ and $$\overrightarrow r .\overrightarrow a = 0,$$ then the value of $$\overrightarrow r .\overrightarrow b $$ is

If $$\overrightarrow a $$ and $$\overrightarrow b $$ are vectors in space given by $$\overrightarrow a = {{\widehat i - 2\widehat j} \over {\sqrt 5 }}$$ and $$\overrightarrow b = {{2\widehat i + \widehat j + 3\widehat k} \over {\sqrt {14} }},$$ then find the value of $$\,\left( {2\overrightarrow a + \overrightarrow b } \right).\left[ {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow a - 2\overrightarrow b } \right)} \right].$$

If the distance between the plane $$Ax-2y+z=d$$ and the plane containing the lines $${{x - 1} \over 2} = {{y - 2} \over 3} = {{z - 3} \over 4}$$ and $${{x - 2} \over 3} = {{y - 3} \over 4} = {{z - 4} \over 5}\,$$ is $$\sqrt 6 \,\,,$$ then find $$\left| d \right|.$$