The volume enclosed by the plane $|3x-4|+|2y-3|+|z+4|=3$ is
Try: $$\bigg|x-\frac{4}{3}\bigg|+\frac{2}{3}\bigg|y-\frac{3}{2}\bigg|+\frac{1}{3}\bigg|z+4\bigg|=1$$
Put $\displaystyle x-\frac{4}{3}=X,y-\frac{3}{2}=Y,z+4=Z$
So $$|X|+\frac{|Y|}{\frac{3}{2}}+\frac{|Z|}{3}=1.$$
Could some helpe to solve it. Thanks