find the angle between the line`(x+1)/2=y/3=(z-3)/6 ` and the plane is `10x+2y-11z=3`

Explanation

Let the given planes intersect on the line with direction ratios l, m and n. In that case.
`(2+lamda)(l)/(a)+(1-2lamda)(m)/(b)+(2-lamda)(n)/(c)=0" "(i)`
and `(4l)/(a)-(3-5mu)(m)/(b)+4mu.(n)/(c)=0" "(ii)`
Hence, `(l//a)/(6-6mu-3lamda-3lamdamu)=(m//b)/(8-8mu-4lamda=4lamdamu)`
`=(n//c)/(-10+10mu+5lamda+5lamdamu)`
or `(l//a)/(3(2-2mu-lamda-lamdamu))=(m//b)/(4(2-2mu=lamda-lamdamu))`
`=(n//c)/((-5(2-2mu-lamda-lamdamu))`
`(l//a)/(3)=(m//b)/4=(n//c)/(-5)" "("provided " 2-2mu-lamda-lamdamune0)`
which are independent of `lamdaand mu`. Hence, a line with driection ratios (3a, 4b, -5c) lies in both the planes.
For `2-2mu-lamda-lamdamu=0orlamda=(2(l-mu))/(1+mu)`, planes (i)
and (ii) coincide with each other. Hence, the two given familise of planes intersect on the same line.