For the truth table above, abc is the input and xyz is the output.
I do know how to draw the diagram using a decoder in case of when I have only one output. How can I draw the diagram using a decoder when I have 3 outputs for this case?
it looks like you nearly have it.
X is high if the input decodes to the '100 value, Y is high if it decodes to '010, '011, '101 or '110 and Z is high if it decodes to '001 '011 '101 or '111.
There is one error in your drawing: X, Y and Z are already your outputs. You don't need to "or" them together at the end.
Then take a look at the drawing and see if you can understand how the ABC decoders are redundant and how you can reuse the outputs from just one decoder to compute X, Y and Z so you only need one decoder.