Given: 

Equation of planes: $x+y+z=6$ and $2x+3y+4z+5=0$

The given equations can be re-written as: 

$x+y+z-6=0$ and $2x+3y+4z+5=0$

As we know that, the equation of a plane passing through the line of intersection of two planes $a_{1}x+b_{1}y+c_{1}z+d_1=0$ and $a_{2}x+b_{2}y+c_{2}z+d_2=0$ is given by:

$(a_{1}x+b_{1}y+c_{1}z+d_1)+\lambda (a_{2}x+b_{2}y+c_{2}z+d_2)=0$ where $\lambda \in R$

$(x+y+z-6)+\lambda (2x+3y+4z+5)=0$ ...(1)

$\because$ It is given that plane passing through the line line of intersection of the plane $x+y+$ $z=6$ and $2x+3y+4z+5=0$ also passes through the point (1,1,1).

i.e The point (1,1,1) will satisfy the equation (1)

$(1+1+1-6)+\lambda (2+3+4+5)=0$

$-3+14\lambda =0$

$\lambda =\frac{3}{14}$

Now by substituting $\lambda =\frac{3}{14}$ in equation (1) we get

$(x+y+z-6)+\frac{3}{14}.(2x+3y+4z+5)=0$

$20x+23y+26z-69=0$

$\begin{array}{rl}& \text{ Given, }P=\left[\begin{array}{ll}1& 0\\ \frac{1}{2}& 1\end{array}\right]\\ & \Rightarrow P^2=\left[\begin{array}{ll}1& 0\\ \frac{1}{2}& 1\end{array}\right]\left[\begin{array}{ll}1& 0\\ \frac{1}{2}& 1\end{array}\right]=\left[\begin{array}{ll}1& 0\\ 1& 1\end{array}\right]\\ & \Rightarrow P^3=\left[\begin{array}{ll}1& 0\\ 1& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ \frac{1}{2}& 1\end{array}\right]=\left[\begin{array}{ll}1& 0\\ \frac{3}{2}& 1\end{array}\right]\\ & \Rightarrow P^4=\left[\begin{array}{ll}1& 0\\ 1& 1\end{array}\right]\left[\begin{array}{ll}1& 0\\ 1& 1\end{array}\right]=\left[\begin{array}{ll}1& 0\\ 2& 1\end{array}\right]\\ & ⋮\\ & \Rightarrow P^n=\left[\begin{array}{cc}1& 0\\ \frac{n}{2}& 1\end{array}\right]\\ & \text{ Hence, }{P}^{50}=\left[\begin{array}{cc}1& 0\\ 25& 1\end{array}\right]\end{array}$

Given,

$P(n)=n^2+n$

By putting $n=1$, we get

$P(1)=1^2+1$

$=2$

As we can say that $2$ is an even number.

Let $P(k)$ is true for $n=k$.

$P(k)=(k^2+k)$ will be even.

$(k^2+k)=2m$ for some natural number $m\dots (i)$

By putting $n=k+1$, we get

$P(k+1)=(k+1{)}^2+(k+1)$

$=k^2+3k+2$

$=(k^2+k)+2(k+1)$

Using equation $(i)$, we get

$P(k+1)=2m+2(k+1)$

$=2[m+(k+1)]$, which is even.

So, $P(k+1)$ is even.

$\Rightarrow P(k+1)$ is true, whenever $P(k)$ is true.

Thus, $P(1)$ is true and $P(k+1)$ is true, whenever $P(k)$ is true.

Thus, by the Principle of Mathematical Induction, $P(n)$ is true for all $n\in N$ i.e., $p(n)=(n^2+n)$ is even.

Given,

$\cos \left({3015}^{\circ }\right)$

$=\cos ({360}^{\circ }\times 8^{\circ }+{135}^{\circ })$$(\because \cos (2n\pi \pm \theta )=\cos \theta )$

$=\cos \left({135}^{\circ }\right)$

$=\cos ({90}^{\circ }+{45}^{\circ })$$(\because \cos (90+\theta )=-\sin \theta )$

$=-\sin {45}^{\circ }$

$=\frac{-1}{\sqrt{2}}$

Let $S$ be the sample space

Number of all combinations of $n$ things, taken $r$ at a time, is given by ${}^n{C}_r=\frac{n!}{(r)!(n-r)!}$

Then, $n(S)={}^{52}{C}_2$

$=\frac{52\times 51}{(2\times 1)}$

$=1326$

Let $E=$ event of getting $2$ kings out of $4$

$\therefore n(E)={}^4{C}_2=\frac{4\times 3}{2\times 1}=6$

$\therefore P(E)=\frac{n(E)}{n(S)}$

$=\frac{6}{1326}$

$=\frac{1}{221}$

Given,

Slope of line = -2

Lines are:

$2x-y=1$....(i)

$x+2y=3$...(ii)

On multiplying equation (i) by 2 we get,

$4x-2y=2$....(iii)

Adding equation (ii) and (iii) we get,

$5x=5$

$\Rightarrow x=1$

On putting value of $x$ in equation (i),

$2(1)-y=1$

$\Rightarrow y=1$

Therefore,

The intersection point is $(1,1)$

So line has the slope m = $-2$ and passes through $(1,1)$

Therefore,

Equation of the perpendicular line is:

$(y-y_1)=m(x-x_1)$

$\Rightarrow y-1=-2(x-1)$

$\Rightarrow y+2x-3=0$

Let $\mathrm{\Delta }=\left|\begin{array}{lll}1-a& a-b-c& b+c\\ 1-b& b-c-a& c+a\\ 1-c& c-a-b& a+b\end{array}\right|$

Applying ${\mathrm{C}}_2\to {\mathrm{C}}_2+{\mathrm{C}}_3,$ we get:

$\mathrm{\Delta }=\left|\begin{array}{lll}1-a& a& b+c\\ 1-b& b& c+a\\ 1-c& c& a+b\end{array}\right|$

Applying ${\mathrm{C}}_1\to {\mathrm{C}}_1+{\mathrm{C}}_2,{\mathrm{C}}_3\to {\mathrm{C}}_3+{\mathrm{C}}_2$, we get:

$\mathrm{\Delta }=\left|\begin{array}{lll}1& a& a+b+c\\ 1& b& b+c+a\\ 1& c& c+a+b\end{array}\right|$

Taking common $\mathrm{a}+\mathrm{b}+\mathrm{c}$ from ${\mathrm{C}}_3,$ we get:

$\mathrm{\Delta }=(a+b+c)\left|\begin{array}{lll}1& a& 1\\ 1& b& 1\\ 1& c& 1\end{array}\right|$

We know that if two columns of a determinant are identical, the value of the determinant is zero.

$\therefore \mathrm{\Delta }=0$