Subtracting (ii) from (i),we have −2y−z=−5⇒2y+z=5....(iv) Multiply (ii) by 2 and subtracting (iii) from it, we obtain 5y−8z=2.....(v) Multiply (iv) by 8 and adding (v) to it, we have 21y=42⇒y=2.....(v) Substituting y=2 in (iv), we get 2×2+z=5⇒z=5−4=1 Substituting these values in (i),we get x+2−3=0⇒x=3−2=1 Hence, the required vector is