$\frac{{x - 0}}{3} = \frac{{y - 0}}{1} = r$
For foot of perpendicular
$r = \frac{{ - \left( {\left( {3 \times 0} \right) + \left( {1 \times 0} \right) - \lambda } \right)}}{{{3^2} + {1^2}}} = \frac{\lambda }{{10}}$
So, food of perpendicular $P = \left( {\frac{{3\lambda }}{{10}},\frac{\lambda }{{10}}} \right)$
Given the line meets $X$-asix at $A = \left( {\frac{\lambda }{3},0} \right)$
and meets $Y$-axis at $B = \left( {o,\lambda } \right)$
So,
$BP = \sqrt {{{\left( {\frac{{3\lambda }}{{10}}} \right)}^2} + {{\left( {\frac{\lambda }{{10}} - \lambda } \right)}^2}} \Rightarrow BP = \sqrt {\frac{{9{\lambda ^2}}}{{100}} + \frac{{81{\lambda ^2}}}{{100}}} $
$ \Rightarrow BP = \sqrt {\frac{{90{\lambda ^2}}}{{100}}} $
Now,$PA = \sqrt {{{\left( {\frac{\lambda }{3} - \frac{{3\lambda }}{{10}}} \right)}^2} + {{\left( {0 + \frac{\lambda }{{10}}} \right)}^2}} $
$PA = \sqrt {\frac{{{\lambda ^2}}}{{900}} + \frac{{{\lambda ^2}}}{{100}}} \Rightarrow PA = \sqrt {\frac{{10{\lambda ^2}}}{{900}}} $
Therefore $BP:PA = 3:1$
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$1.$ the selection of four red marbles.
$2.$ the selection of one white and three red marbles.
$3.$ the selection of one white, one blue and two red marbles.
$4.$ the selection of one marble of each colour.
The smallest total number of marbles satisfying the given condition is
(where $S = p[{p^4} - 5{p^2}q + 5{q^2}],\;P = {p^2}{q^2}$$({p^4} - 5{p^2}q + 4{q^2})$