Perpendiculars are drawn from points on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$ to the plane $x+y+$ $z=3$. The foot of perpendiculars lie on the line

For $$a > b > c > 0,$$ the distance between $$(1, 1)$$ and the point of intersection of the lines $$ax + by + c = 0$$ and $$bx + ay + c = 0$$ is less than $$\left( {2\sqrt 2 } \right)$$. Then

A vertical line passing through the point $$(h,0)$$ intersects the ellipse $${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1$$ at the points $$P$$ and $$Q$$. Let the tangents to the ellipse at $$P$$ and $$Q$$ meet at the point $$R$$. If $$\Delta \left( h \right)$$$$=$$ area of the triangle $$PQR$$, $${{\Delta _1}}$$ $$ = \mathop {\max }\limits_{1/2 \le h \le 1} \Delta \left( h \right)$$ and $${{\Delta _2}}$$ $$ = \mathop {\min }\limits_{1/2 \le h \le 1} \Delta \left( h \right)$$, then $${8 \over {\sqrt 5 }}{\Delta _1} - 8{\Delta _2} = $$

The value of $$\cot \left( {\sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}} \left( {1 + \sum\limits_{k = 1}^n {2k} } \right)} \right)$$ is