The correct option is
B ¯r.(4^i−3^j+2^k)=9Consider the problem,
As we know that equation of line passing through (x1,y1,z1)
a(x−x1)+b(y−y1)+z(z−z1)=0
where a,b,c are direction ratios
Now, equation of plane passing through (2,5,−3)
a(x−2)+b(y−5)+c(z+3)=0 ---- 1
Since it also passes through (−2,−3,5)
a(−2−2)+b(−3−5)+c(5+3)=0
−4a−8b+8c=0
a+2b−2c=0 ----- (i)
And it also passes through (5,3,−3)
Therefore,
a(5−2)+b(3−5)+c(−3+3)=0
3a−2b=0 ----- (ii)
Solving (i) and (ii) by cross multiplication we get
a0+2=b−6−0=c−2−6
a2=b−6=c−8
a−2=b6=c8=k Consider
Therefore,
a=−2k,b=6k,c=8k
Now, from equation 1
a(x−2)+b(y−5)+c(z+3)=0
−2k(x−2)+6k(y−5)+8k(z+3)=0
−2(x−2)+6(y−5)+8(z+3)=0
−2x+4+6y−30+8z+24=0
x−3y−4z+1=0
Let →r be the position vector. which is x^i+y^j+z^k
→r(^i−3^j−4^k)=−1