The equation of the plane passing through the point $(1,1,1)$ and through the line of intersection of $x+2y-z+1=0$ and $3x-y-4z+3=0$ is

Find the equation of the plane passing through the intersection of the planes $x + 2y + 3z = 2$ and $x - y + z = 3$ and at a distance of $\frac{2}{\sqrt{3}}$ from the point $(3, 1, -1)$.

If the plane passing through the points $\hat{i}+\hat{j}+\hat{k}$, $2\hat{i}-\hat{k}$ and the origin meets the line passing through the points $\hat{i}+3\hat{j}-2\hat{k}$ and $\hat{i}-\hat{j}+3\hat{k}$ at the point $A$, then $A=$

$X$ intercept of the plane containing the line of intersection of the planes $x-2y+z+2=0$ and $3x-y-z+1=0$ and also passing through $(1,1,1)$ is

Let $P (-2,-1,1)$ and $Q \left(\frac{56}{17}, \frac{43}{17}, \frac{111}{17}\right)$ be the vertices of the rhombus $PRQS$. If the direction ratios of the diagonal $RS$ are $\alpha, -1, \beta$,where both $\alpha$ and $\beta$ are integers of minimum absolute values,then $\alpha^{2}+\beta^{2}$ is equal to $.....$

If the three planes $x = 5$,$2x - 5ay + 3z - 2 = 0$,and $3bx + y - 3z = 0$ pass through a common line,then the value of $(a, b)$ is: