Q.D and E divide sides BC and CA of a triangle ABC in the ratio $2:3$ respectively. Find the position vector of the point of intersection of AD and BE and the ratio in which this point divides AD and BE.

Let $\bar a,\bar b,\bar c$ be position vectors of $A,B,C$. Since $D$ divides $BC$ in ratio $2:3$,

$\bar d=\dfrac{2\bar c+3\bar b}{5}$; since $E$ divides $CA$ in ratio $2:3$, $\bar e=\dfrac{2\bar a+3\bar c}{5}$.

Let $P$ divide $AD$ in ratio $m:1$: $\bar p=\dfrac{\bar a+m\bar d}{1+m}=\dfrac{5\bar a+m(3\bar b+2\bar c)}{5(1+m)}$.

Let $P$ divide $BE$ in ratio $k:1$: $\bar p=\dfrac{\bar b+k\bar e}{1+k}=\dfrac{5\bar b+k(2\bar a+3\bar c)}{5(1+k)}$.

Matching the coefficients of $\bar a,\bar b,\bar c$ across the two expressions for $\bar p$ (since $\bar a,\bar b,\bar c$ are position vectors of a genuine, non-degenerate triangle) gives three equations; solving them …

Express D and E via the section formula, write P's position vector two ways (as a point on AD and as a point on BE) using unknown ratios m:1 and k: …

Equating coefficients incorrectly when the two expressions are not first brought to a common denomina …