The distance of the point (1, –2, 3) from

the plane x – y + z = 5 measured parallel to

the line $${x \over 2} = {y \over 3} = {z \over { - 6}}$$ is :

Suppose the vectors x₁, x₂ and x₃ are the
solutions of the system of linear equations,
Ax = b when the vector b on the right side is equal to b₁, b₂ and b₃ respectively. if

$${x_1} = \left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right]$$, $${x_2} = \left[ {\matrix{ 0 \cr 2 \cr 1 \cr } } \right]$$, $${x_3} = \left[ {\matrix{ 0 \cr 0 \cr 1 \cr } } \right]$$

$${b_1} = \left[ {\matrix{ 1 \cr 0 \cr 0 \cr } } \right]$$, $${b_2} = \left[ {\matrix{ 0 \cr 2 \cr 0 \cr } } \right]$$ and $${b_3} = \left[ {\matrix{ 0 \cr 0 \cr 2 \cr } } \right]$$,
then the determinant of A is equal to :

If a and b are real numbers such that
$${\left( {2 + \alpha } \right)^4} = a + b\alpha $$
where $$\alpha = {{ - 1 + i\sqrt 3 } \over 2}$$ then a + b is equal to :

If the perpendicular bisector of the line segment joining the points P(1 ,4) and Q(k, 3) has y-intercept equal to –4, then a value of k is :