x−2−1−3y−1−20z−10−2=0 Solve the above determinat then, 2x−y−3z=0 The above equation can be expressed in vector form as, r⋅(2i^−j^−3k^)=0 If e is unit vector perpendicular to given plane then, e=22+(−1)2+(−3)22i^−j^−3k^=141(2i^−j^−3k^) And, a=2i^−3j^+6k^ The projection of a on e is calculated as, a⋅e=144+143−1418=14−11 The projection vector of a on e is calculated as, a⋅e=−1411(141(2i^−j^−3k^))=1411(−2i+j^+3k)