Given equation of plane is
2x−3y+4z−6=0
⇒2x−3y+4z=6
Let point
P(x1,y1,z1) be foot of perpendicular from origin.
Direction ratios of the normal vector to the plane are
2,−3,4
Direction ratios of vector
−−→OP is
x1,y1,z1
As vector
−−→OP is parallel to normal vector
→n, so the direction ratios are proportional,
x12=y1−3=z14=k
⇒x1=2k,y1=−3k,z1=4k
As this point lies on the plane, so
2x1−3y1+4z1−6=0
⇒2(2k)−3(−3k)+4(4k)−6=0
⇒4k+9k+16k=6
⇒k=629
Therefore,
x1=2k=1229
y1=−3k=−1829
z1=4k=2429
Hence, the required coordinate of the foot of perpendicular are
(1229,−1829,2429)