A plane is making intercepts 2, 3, 4 on X, Y | vedclass

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(C) The equation of the plane with intercepts $a=2, b=3, c=4$ is given by $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$. Multiplying by $12$,we get $6

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$\cos ^{-1} \sqrt{\frac{11}{61}}$

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(C) The equation of the plane with intercepts $a=2, b=3, c=4$ is given by $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$. Multiplying by $12$,we get $6x + 4y + 3z = 12$. The normal vector to this plane is $\vec{n_1} = (6, 4, 3)$.The second plane is perpendicular to the line joining $B(1, 2, 3)$ and $C(-2, 3, 4)$. The direction ratios of the line $BC$ are $(-2-1, 3-2, 4-3) = (-3, 1, 1)$. Since the plane is perpendicular to this line,the normal vector to the second plane is $\vec{n_2} = (-3, 1, 1)$.The equation of the second plane passing through $(-1, 6, 2)$ is $-3(x+1) + 1(y-6) + 1(z-2) = 0$,which simplifies to $-3x + y + z - 11 = 0$.The angle $	heta$ between the two planes is given by $\cos 	heta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.$\vec{n_1} \cdot \vec{n_2} = (6)(-3) + (4)(1) + (3)(1) = -18 + 4 + 3 = -11$.$|\vec{n_1}| = \sqrt{6^2 + 4^2 + 3^2} = \sqrt{36 + 16 + 9} = \sqrt{61}$.$|\vec{n_2}| = \sqrt{(-3)^2 + 1^2 + 1^2} = \sqrt{9 + 1 + 1} = \sqrt{11}$.Thus,$\cos 	heta = \frac{|-11|}{\sqrt{61} \sqrt{11}} = \frac{11}{\sqrt{61} \sqrt{11}} = \frac{\sqrt{11}}{\sqrt{61}} = \sqrt{\frac{11}{61}}$.Therefore,$	heta = \cos ^{-1} \sqrt{\frac{11}{61}}$.

$A$ plane is making intercepts $2, 3, 4$ on $X, Y$ and $Z$-axes respectively. Another plane is passing through the point $(-1, 6, 2)$ and is perpendicular to the line joining the points $(1, 2, 3)$ and $(-2, 3, 4)$. Then the angle between the two planes is

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