The angle between the line `vecr.(hati+hatj-3hatk)+lambda(2hati+2hatj+hatk)` and the plane `vecr.(6hati-3hatj+2hatk)=5`, is

Explanation

We know that the angle `phi` between the line `vecr=veca+lambdavecb` and the plane `vecr.vecn=q` is given by
`sinphi=|vecb.vecn|/(|vecb||vecn|)`.
Here, `vecb=(2hati+2hatj+hatk)` and `vecn=(6hati-3hatj+2hatk)`.
`therefore sinphi=|(2hati+2hatj+hatk).(6hati-3hatj+2hatk)|/(sqrt(2^(2)+2^(2)+1^(2))sqrt(6^(2)+(-3)^(2)+2^(2))`
`=|(12-6+2)|/(sqrt(9) xx sqrt(49)) = 8/(3 xx 7)=8/21`.
`rArr phi=sin^(-1)(8/21)`.
Hence, the angle between the given line and the given plane is `sin^(-1)(8/21)`.