You should remember how to write the vector when the problem provides the coordinates of the initial point and terminal point of vector such that:

`bar(OA) = (x_A - x_O)*bar i + (y_A - y_O)*bar j`

`bar(OA) = (-2- 0)*bar i + (3 - 0)*bar j`

`bar(OA) = -2bar i + 3bar j`

Multiplying `bar(OA) ` by 2 yields:

`2bar(OA) = 2(-2bar i + 3bar j) `

`2bar(OA) = -4bar i + 6bar j`

`bar(OB) = (x_B - x_O)*bar i + (y_B - y_O)*bar j `

`bar(OB) = (4-0)*bar i + (2-0)*bar j`

`bar(OB) = 4bar i + 2bar j`

Multiplying `bar(OB)` by 3 yields:

`3bar(OB) = 3(4bar i + 2bar j) `

`3bar(OB) = 12bar i + 6bar j`

Adding `2bar(OA)` and `3bar(OB)` yields:

`2bar(OA)+ 3bar(OB) =-4bar i + 6bar j + 12bar i + 6bar j` Collecting like terms yields:

`2bar(OA) + 3bar(OB) = 8bar i + 12bar j`

The problem provides the information that `2bar(OA) + 3bar(OB) = bar(OC)` such that:

`bar(OC) = 8bar i + 12bar j`

`bar(OC) = (x_C - x_O)*bar i + (y_C - y_O)*bar j`

`bar(OC) = x_C *bar i + y_C*bar j`

Comparing `bar(OC) = x_C *bar i + y_C*bar j` to `bar(OC) = 8bar i + 12bar j` yields `x_C = 8` and `y_C = 12`

**Hence, evaluating the coordinate of the point C yields C(8,12).**