Let the vectors $$\overrightarrow {PQ} ,\,\,\overrightarrow {QR} ,\,\,\overrightarrow {RS} ,\,\,\overrightarrow {ST} ,\,\,\overrightarrow {TU} ,$$ and $$\overrightarrow {UP} ,$$ represent the sides of a regular hexagon.

STATEMENT-1: $$\overrightarrow {PQ} \times \left( {\overrightarrow {RS} + \overrightarrow {ST} } \right) \ne \overrightarrow 0 .$$ because
STATEMENT-2: $$\overrightarrow {PQ} \times \overrightarrow {RS} = \overrightarrow 0 $$ and $$\overrightarrow {PQ} \times \overrightarrow {ST} \ne \overrightarrow 0 \,\,.$$

Consider the planes $$3x-6y-2z=15$$ and $$2x+y-2z=5.$$

STATEMENT-1: The parametric equations of the line of intersection of the given planes are $$x=3+14t,y=1+2t,z=15t.$$ because

STATEMENT-2: The vector $${14\widehat i + 2\widehat j + 15\widehat k}$$ is parallel to the line of intersection of given planes.

Let $$\,{V_r}$$ denote the sum of first r terms of an arithmetic progression (A.P.) whose first term is r and the common difference is (2r-1). Let $${T_r} = \,{V_{r + 1}} - \,{V_r} - 2\,\,\,and\,\,\,{Q_r} = \,{T_{r + 1}} - \,{T_r}\,for\,r = 1,2,...$$

$${T_r}$$ is always

A man walks a distance of 3 units from the origin towards the north-east ($$N\,{45^ \circ E }$$) direction. From there, he walks a distance of 4 units towards the north-west $$\left( {N\,{{45}^ \circ }\,W} \right)$$ direction to reach a point P. Then the position of P in the Argand plane is