$A$ 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

The plane $3x + 4y + 6z + 7 = 0$ is rotated about the line $r = (\hat{i} + 2\hat{j} - 3\hat{k}) + t(2\hat{i} - 3\hat{j} + \hat{k})$ until the plane passes through the origin. The equation of the plane in the new position is

If $\lambda_1 < \lambda_2$ are two values of $\lambda$ such that the angle between the planes $P_1: \vec{r} \cdot (3 \hat{i} - 5 \hat{j} + \hat{k}) = 7$ and $P_2: \vec{r} \cdot (\lambda \hat{i} + \hat{j} - 3 \hat{k}) = 9$ is $\sin^{-1}\left(\frac{2 \sqrt{6}}{5}\right)$,then the square of the length of the perpendicular from the point $(38 \lambda_1, 10 \lambda_2, 2)$ to the plane $P_1$ is $...........$.

The equation of the line,passing through $(1, 2, 3)$ and parallel to the planes $x - y + 2z = 5$ and $3x + y + z = 6$,is

The line of intersection of the planes $\vec{r} \cdot (3\hat{i} - \hat{j} + \hat{k}) = 1$ and $\vec{r} \cdot (\hat{i} + 4\hat{j} - 2\hat{k}) = 2$ is parallel to which vector?

Let $L$ be the line of intersection of the planes $2x + 3y + z = 1$ and $x + 3y + 2z = 2$. If $L$ makes an angle $\alpha$ with the positive $X$-axis,then $\cos \alpha$ equals