The distance from the origin to the image of $(1,1)$ with respect to the line $x+y+5=0$ is (in $\sqrt{2}$)

If the foot of the perpendicular drawn from the origin to a straight line is $(3, -4)$,then the equation of the line is:

The perpendicular bisector of the line segment joining $P(1, 4)$ and $Q(k, 3)$ has a $y-$intercept of $-4$. Then a possible value of $k$ is:

Find the image of the point $(-1, 3)$ with respect to the line $x - y + 1 = 0$.

The equation of the straight line passing through the point $(a \cos^3 \theta, a \sin^3 \theta)$ and perpendicular to the line $x \sec \theta + y \csc \theta = a$ is:

Consider the lines $x(3 \lambda+1)+y(7 \lambda+2)=17 \lambda+5$,where $\lambda$ is a parameter. All these lines pass through a fixed point $P$. One of these lines (say $L$) is farthest from the origin. If the distance of $L$ from the point $(3,6)$ is $d$,then the value of $d^2$ is