#### Answer

$-3i+(2^{5/2}-2)j+2k$

#### Work Step by Step

We need to integrate the given integral as follows:
$[\int_1^2 (6-6t)dt]i+[\int_1^2 3 \sqrt t dt] j+[\int_0^2 (4/t^2)dt]= (6t-3t^2)_1^2i+(2t^{3/2})_1^2j+(\dfrac{-4}{t})_1^2k$
or,
$(6t-3t^2)_1^2i+(2t^{3/2})_1^2j+(\dfrac{-4}{t})_1^2k=(0-3)i+(2^{5/2}-2)j+[-2-(-4)] k$
Thus,
$(0-3)i+(2^{5/2}-2)j+[-2-(-4)] k=-3i+(2^{5/2}-2)j+2k$