Let the cost of 1 kg onion, wheat, and rice be $x$, $y$, and $z$ respectively. The system of equations is:

$$4x + 3y + 2z = 60$$ $$2x + 4y + 6z = 90$$ $$6x + 2y + 3z = 70$$

This system can be written in the form $AX = B$, where $A = \begin{bmatrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, and $B = \begin{bmatrix} 60 \\ 90 \\ 70 \end{bmatrix}$.

To find $X$, we calculate $A^{-1}$. First, we find the determinant of $A$:

$$|A| = 4(12-12) - 3(6-36) + 2(4-24) = 0 + 90 - 40 = 50$$

Since $|A| \neq 0$, the inverse exists.

Next, we find the cofactors of matrix $A$ to determine $\text{adj}(A)$.

$$\text{adj}(A) = \begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix}$$

Thus, $A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{50} \begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix}$.

Finally, we calculate $X = A^{-1}B$:

$$X = \frac{1}{50} \begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix} \begin{bmatrix} 60 \\ 90 \\ 70 \end{bmatrix} = \frac{1}{50} \begin{bmatrix} 250 \\ 400 \\ 400 \end{bmatrix} = \begin{bmatrix} 5 \\ 8 \\ 8 \end{bmatrix}$$

Hence, the cost of onion is ₹5/kg, wheat is ₹8/kg, and rice is ₹8/kg.