Prove that there is only one common intersecting line of planes `vecr . (3i +4j +2k) = 0, vecr . (3i - j +3k) = 2` and `vecr . (5j - k) + 2 =0`

The line of intersection of the planes vec r . (3i-j+k) =1 and vec r . (i+4j-2k) = 2 is parallel to the vector

Line of intersection of the two planes barr.(3i - j + k)=1 and barr.(i+4j-2k)=2 is parallel to the vector

Find the equation of the plane through the intersection of the planes vecr.(hat(i) + 3hat(j)) - 6 = 0 and vecr.(3hat(i) - hat(j) - 4hat(k)) = 0 , whose perpendicular distance from origin is unity.

The vectors 2i - 3j + k, I - 2j + 3k, 3i + j - 2k

The distance between the planes r. (i+ 2j - 2k ) + 5 =0 and r. (2i + 4j - 4k ) - 16 =0 is

Find the length of the perpendicular from origin to the plane vecr.(3i-4j-12k)+39=0 .

Find the vector equation of the plane passing through the intersection of the planes vecr.(hat(i) + hat(j) + hat(k)) = 6 and vecr.(2hat(i) + 3hat(j) + 4hat(k))= - 5 and the point (1, 1, 1).

If a = 3i - 2j + k, b = 2i - 4j - 3k, c = -i + 2j + 2k then a + b + c =