If the equation of the plane passing through the line of intersection of the planes $$2 x-y+z=3,4 x-3 y+5 z+9=0$$ and parallel to the line $$\frac{x+1}{-2}=\frac{y+3}{4}=\frac{z-2}{5}$$ is $$a x+b y+c z+6=0$$, then $$a+b+c$$ is equal to :

One vertex of a rectangular parallelopiped is at the origin $$\mathrm{O}$$ and the lengths of its edges along $$x, y$$ and $$z$$ axes are $$3,4$$ and $$5$$ units respectively. Let $$\mathrm{P}$$ be the vertex $$(3,4,5)$$. Then the shortest distance between the diagonal OP and an edge parallel to $$\mathrm{z}$$ axis, not passing through $$\mathrm{O}$$ or $$\mathrm{P}$$ is :

Let the plane P pass through the intersection of the planes $$2x+3y-z=2$$ and $$x+2y+3z=6$$, and be perpendicular to the plane $$2x+y-z+1=0$$. If d is the distance of P from the point ($$-$$7, 1, 1), then $$\mathrm{d^{2}}$$ is equal to :

The shortest distance between the lines

$${{x - 5} \over 1} = {{y - 2} \over 2} = {{z - 4} \over { - 3}}$$ and

$${{x + 3} \over 1} = {{y + 5} \over 4} = {{z - 1} \over { - 5}}$$ is :