The vertices B and C of a $$\Delta $$ABC lie on the line,

$${{x + 2} \over 3} = {{y - 1} \over 0} = {z \over 4}$$ such that BC = 5 units.

Then the area (in sq. units) of this triangle, given that the point A(1, –1, 2), is :

The total number of matrices
$$A = \left( {\matrix{ 0 & {2y} & 1 \cr {2x} & y & { - 1} \cr {2x} & { - y} & 1 \cr } } \right)$$
(x, y $$ \in $$ R,x $$ \ne $$ y) for which A^TA = 3I₃ is :-

If $$f(x) = [x] - \left[ {{x \over 4}} \right]$$ ,x $$ \in $$ 4 , where [x] denotes the greatest integer function, then

If the function $$f(x) = \left\{ {\matrix{ {a|\pi - x| + 1,x \le 5} \cr {b|x - \pi | + 3,x > 5} \cr } } \right.$$
is continuous at x = 5, then the value of a – b is :-