Let $\alpha$ be the common root of the equation $3 x^2-7 x+2=0$ and $k x^2+7 x-3=0$, so
$3 \alpha^2-7 \alpha+2=0$ $\ldots$ (i)
and $\quad k \alpha^2+7 \alpha-3=0$ $\ldots$ (ii)
By cross multiplication, we get
$\frac{\alpha^2}{21-14}=\frac{-\alpha}{-9-2 k}=\frac{1}{21+7 k}$
$\Rightarrow \quad \frac{\alpha^2}{7}=\frac{\alpha}{2 k+9}=\frac{1}{7(k+3)}$
$\therefore \quad \frac{1}{k+3}=\frac{(2 k+9)^2}{49(k+3)^2}$
$\because k$ is positive, so
$49(k+3)=4 k^2+36 k+81$
$\Rightarrow \quad 4 k^2-13 k-66=0$
$\Rightarrow \quad 4 k^2-24 k+11 k-66=0$
$\Rightarrow \quad 4 k(k-6)+11(k-6)=0$
$\Rightarrow \quad k=6 \quad[\because k$ is positive $]$