Explanation

(a)
Let $\alpha$ be the common roots to the equations
$x^{2} - kx - 21 = 0$ and $x^{2} - 3kx + 35 = 0$
$\therefore\ \ \ \alpha^{2} - k\alpha - 21 = 0$ and $\alpha^{2} - 3k\alpha + 35 = 0$
Now, by cross multiplication method, we get
$\frac{\alpha^{2}}{( - 35k - 63k)} = \frac{\alpha}{( - 21 - 35)} = \frac{1}{( - 3k + k)}$
$\Rightarrow \ \ \frac{\alpha^{2}}{- 98k} = \frac{\alpha}{- 56} = \frac{1}{- 2k}$
$\Rightarrow \ \ \ \ \frac{\alpha^{2}}{- 98k} = \frac{1}{- 2k}\ \ \Rightarrow \ \ \ \alpha^{2} = 49$ ...(i)
And $\frac{\alpha}{- 56} = - \frac{1}{2k}\ \ \Rightarrow \ \ \ \alpha = \frac{28}{k}$ ...(ii)
From Eqs. (i) and (ii),
$\frac{28 \times 28}{k^{2}} = 49$
$\Rightarrow \ \ \ k^{2} = 16$
$\Rightarrow \ \ k = \pm 4$