The correct option is
A True
A(4,5,1),B(5,3,4);C(4,1,6) & D(3,3,3)
→AB=(5−4)^i+(3−5)^j+(4−1)^k
=^i−2^j+3^k
→AC=(4−4)^i+(1−5)^j+(6−1)^k
=−4^j+5^k
→AD=(3−4)^i+(3−5)^j+(3−1)^k
=−^i−2^j+2^k
∴|→AB,→AC,→AD|=∣∣
∣∣1−230−45−1−22∣∣
∣∣
=1(−8+10)+2((2×0)+5)+3((−2×0)−4)
=2+10−12=0
Hence four points A, B, C, D afe coplanar.