If the position vectors of the vertices A, B, and C of a triangle ABC are respectively 4^i+7^j+8^k, 2^i+3^j+4^k, and 2^i+5^j+7^k, then the position vector of the point where the bisector of ∠A meets BC is:

Let angular bisector of A meets side BC at point P(x, y, z)
By angular bisector theorem we can say that AB:AC=BP:PC
∴BP:PC=c:b
⇒BP:PC=6:3=2:1=m:n
⇒P(x,y,z)=(mx2+nx1m+n,my2+ny1m+n,mz2+nz1m+n)
Here B=(2,3,4)=(x2,y2,z2) and C=(2,5,7)=(x3,y3,z3)
Subtituting values, we get;
P(x,y,z)=((2)(2)+(1)(2)2+1,(2)(5)+(1)(3)2+1,(2)(7)+(1)(4)2+1)
P(x,y,z)=(63,133,183)
∴Position vector of point P=13(6i+13j+18k)
Hence, the correct option is 'B'.