Let M be the foot of perpendicular from P(3,4,5) to the given plane, then PM is normal to the plane. So, its DR's are (1,1,1).
∴ Equation of line PM is ‌
x−3
1
‌=‌
y−4
1
=‌
z−5
1
=k‌ (say) ‌ ⇒‌‌x=k+3,y=k+4,z‌=k+5 Let coordinate of M be (k+3,k+4,k+5) Since, point M lies on a plane x+y+z=9. ∴ It satisfies the equation of plane. ∴‌1⋅(k+3)+1⋅(k+4)+1⋅(k+5)=9 ⇒3k+12=9 ⇒3k=−3 ⇒k=−1 Put k=−1 in Eq. (i), we get The coordinate of M(−1+3,−1+4,−1+5) i.e., ‌‌M(2,3,4)