(3)

Given : x + 2y – 3z = 2

$2x+\lambda y+5z=5$

$14x+3y+\mu z=33$

For infinitely many solutions, we have $∆=0$

$\Rightarrow$$\left|\begin{array}{ccc}1& 2& –3\\ 2& \lambda & 5\\ 14& 3& \mu \end{array}\right|$$=0$

$\Rightarrow 1(\lambda \mu –15)–2(2\mu –70)–3(6–14\lambda )=0$

$\Rightarrow \lambda \mu +42\lambda –4\mu +107=0$

We also have, ${△}_1=2\lambda \mu +99\lambda –10\mu +225$

${△}_2=\mu –13, {△}_3=5\lambda +5$

Now, ${△}_2=0 \mathrm{and} {△}_3=0$

$\therefore 13–\mu =0 \mathrm{and} 5\lambda +5=0 \Rightarrow \mu =13 \Rightarrow \lambda =–1$

Thus, we get $\lambda +\mu =13+(–1)=13–1=12$

(4)

We have, x + y + z = 1

x + 2y +4z = m and x + 4y + 10z = $m^2$

$\therefore △=$$\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 4\\ 1& 4& 10\end{array}\right|$$=1\left(4\right)–1\left(6\right)+1\left(2\right)=0$

For infinite many solutions, $△x=△y=△z=0$

Now, $△x=0 \Rightarrow$$\left|\begin{array}{ccc}1& 1& 1\\ m& 2& 4\\ m^2& 4& 10\end{array}\right|$$=0$

$\Rightarrow 1\left(4\right)–1(10m–4m^2)+1(4m–2m^2)=0$

$\Rightarrow 4–10m+4m^2+4m–2m^2=0$

$\Rightarrow 2m^2–6m+4=0$

$\Rightarrow m^2–3m+2=0 \Rightarrow m=1,2$

$\therefore \alpha =1, \beta =2$

Now, $\sum _{n=1}^{10}(n^{\alpha }+n^{\beta })=\sum _{n=1}^{10}n+\sum _{n=1}^{10}{n}^2$

$=\frac{10\left(11\right)}{2}+\frac{10\left(11\right)\left(21\right)}{6}$ = 55 + 385 = 440.