Let $$f$$ be a real-valued differentiable function on $$R$$ (the set of all real numbers) such that $$f(1)=1$$. If the $$y$$-intercept of the tangent at any point $$P(x,y)$$ on the curve $$y=f(x)$$ is equal to the cube of the abscissa of $$P$$, then find the value of $$f(-3)$$

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 $$\left| d \right|.$$

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 $$\,\left( {2\overrightarrow a + \overrightarrow b } \right).\left[ {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow a - 2\overrightarrow b } \right)} \right].$$

Equation of the plane containing the straight line $${x \over 2} = {y \over 3} = {z \over 4}$$ and perpendicular to the plane containing the straight lines $${x \over 3} = {y \over 4} = {z \over 2}$$ and $${x \over 4} = {y \over 2} = {z \over 3}$$ is