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

$A$ line $L$ is passing through the point $A$ whose position vector is $\hat{i}+2 \hat{j}-3 \hat{k}$ and is parallel to the vector $2 \hat{i}+\hat{j}+2 \hat{k}$. $A$ plane $\pi$ is passing through the points $\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-\hat{j}-\hat{k}$ and is parallel to the vector $\hat{i}-2 \hat{j}$. Then the point where this plane $\pi$ meets the line $L$ is

If the distance of the point $P(43, \alpha, \beta), \beta < 0$, from the line $\vec{r} = 4\hat{i} - \hat{k} + \mu(2\hat{i} + 3\hat{k}), \mu \in R$ along a line with direction ratios $3, -1, 0$ is $13\sqrt{10}$, then $\alpha^{2} + \beta^{2}$ is equal to . . . . . . .

The coordinates of the foot of the perpendicular from the point $(1, -2, 1)$ on the plane containing the lines $\frac{x + 1}{6} = \frac{y - 1}{7} = \frac{z - 3}{8}$ and $\frac{x - 1}{3} = \frac{y - 2}{5} = \frac{z - 3}{7}$ is

The line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3y-\alpha z+\beta=0$,then the value of $\alpha^2+\alpha\beta+\beta^2$ is

Find the equation of the plane passing through the intersection of the planes $x + 2y + 3z - 4 = 0$ and $2x + y - z + 5 = 0$ and perpendicular to the plane $5x + 3y + 6z + 8 = 0$.