A plane which is perpendicular to two planes $$2x - 2y + z = 0$$ and $$x - y + 2z = 4,$$ passes through $$(1, -2, 1).$$ The distance of the plane from the point $$(1, 2, 2)$$ is

Match the following :

Column $$I$$
(A) $$\int\limits_0^{\pi /2} {{{\left( {\sin x} \right)}^{\cos x}}\left( {\cos x\cot x - \log {{\left( {\sin x} \right)}^{\sin x}}} \right)dx} $$
(B) Area bounded by $$ - 4{y^2} = x$$ and $$x - 1 = - 5{y^2}$$
(C) Cosine of the angle of intersection of curves $$y = {3^{x - 1}}\log x$$ and $$y = {x^x} - 1$$ is
(D) Let $${{dy} \over {dx}} = {6 \over {x + y}}$$ where $$y(0)=0$$ then value of $$y$$ when $$x+y=6$$ is

Column $$II$$
(p) $$1$$
(q) $$0$$
(r) $$6\ln 2$$
(s) $${4 \over 3}$$

Let the definite integral be defined by the formula
$$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 2}\left( {f\left( a \right) + f\left( b \right)} \right).} $$ For more accurate result for
$$c \in \left( {a,b} \right),$$ we can use $$\int\limits_a^b {f\left( x \right)dx = \int\limits_a^c {f\left( x \right)dx + \int\limits_c^b {f\left( x \right)dx = F\left( c \right)} } } $$ so
that for $$c = {{a + b} \over 2},$$ we get $$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 4}\left( {f\left( a \right) + f\left( b \right) + 2f\left( c \right)} \right).} $$

$$\int\limits_0^{\pi /2} {\sin x\,dx = } $$

Let the definite integral be defined by the formula
$$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 2}\left( {f\left( a \right) + f\left( b \right)} \right).} $$ For more accurate result for
$$c \in \left( {a,b} \right),$$ we can use $$\int\limits_a^b {f\left( x \right)dx = \int\limits_a^c {f\left( x \right)dx + \int\limits_c^b {f\left( x \right)dx = F\left( c \right)} } } $$ so
that for $$c = {{a + b} \over 2},$$ we get $$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 4}\left( {f\left( a \right) + f\left( b \right) + 2f\left( c \right)} \right).} $$

If $$\mathop {\lim }\limits_{x \to a} {{\int\limits_a^x {f\left( x \right)dx - \left( {{{x - a} \over 2}} \right)\left( {f\left( x \right) + f\left( a \right)} \right)} } \over {{{\left( {x - a} \right)}^3}}} = 0,\,\,$$ then $$f(x)$$ is
of maximum degree