The correct option is A 213(→p+3→q+→p×→q)
Given that |¯p|=2,|¯q|=3,θ=π3⇒¯p.¯q=3
Consider ¯x=λ¯p+μ¯q+t(¯pׯq)
¯pׯx=μ(¯pׯq)+t[(¯p.¯q)¯p−(¯p.¯p)¯q]
¯pׯx=μ(¯pׯq)+t[3¯p−4¯q]
We have,
¯pׯx+2¯q−3¯x=0
⇒μ(¯pׯq)+t[3¯p−4¯q]+2¯q−3(λ¯p−μ¯q+t(¯pׯq))=0
⇒(3t−3λ)¯p+(2−4t−3μ)¯q+(μ−3t)(¯pׯq)=0
p,q,p×q are non coplanar. Hence,
3t−3λ=0,
2−4t−3μ=0
μ−3t=0
On solving we get,
λ=213, t=213 and μ=613
Therefore, ¯x=213[¯p+3¯q+(pׯq)]
Hence, option 'A' is correct.