Given, $X$-intercept $(a)=2$
$$
\begin{aligned}
& Y \text { - intercept }(b)=3 \\
& Z \text { - intercept }(c)=4
\end{aligned}
$$
$\therefore$ Equation of the plane is $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$


$$
\begin{aligned}
& B=(1,2,3) \\
& C=(-2,3,4)
\end{aligned}
$$
$\mathrm{DR}^{\prime} \mathrm{S}$ of $\mathbf{B C}=(-2,-1,3-2,4-3)$
$$
=(-3,1,1)
$$
$\therefore$ Equation of plane passing through $A(-1,6,2)$ and having DR's $(-3,1,1)$ is given by
$$
\begin{aligned}
& a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0 \\
& -3(x+1)+1(y-6)+1(z-2)=0 \\
& -3 x-3+y-6+z-2=0 \\
& -3 x+y+z-11=0
\end{aligned}
$$

Angle between the planes (i) and (ii) is
$$
\begin{aligned}
\cos \theta & =\frac{\left|a_1 a_2+b_1 b_2+c_1 c_2\right|}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}} \\
& =\frac{|(6)(-3)+(4)(1)+(3)(1)|}{\sqrt{36+16+9} \sqrt{9+1+1}} \\
& =\frac{|-18+4+3|}{\sqrt{61} \sqrt{11}} \\
& =\frac{(\sqrt{11})^2}{\sqrt{61} \sqrt{11}} \\
& =\frac{\sqrt{11}}{\sqrt{61}} \\
\cos \theta & =\sqrt{\frac{11}{61}} \\
\therefore \theta & =\cos ^{-1} \sqrt{\frac{11}{61}}
\end{aligned}
$$
$\therefore$ Hence, option (c) is correct.