The sum of the squares of the perpendicular distances of a point $(x, y, z)$ from the coordinate axes is $k$ times the square of the distance of the point from the origin Then, $k=$

Equation of the plane through the mid-point of the line segment joining the points $A(4,5,-10)$ and $B(-1,2,1)$ and perpendicular to $A B$ is

If $\mathbf{r}=(2-\lambda+\mu) \hat{\mathbf{i}}+(1-\mu) \hat{\mathbf{j}}+(2-3 \lambda+2 \mu) \hat{\mathbf{k}}$ is the vector equation of a plane, then the equivalent cartesian equation of the plane is

Let $\pi_1$ be a plane passing through the point $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and perpendicular to the vector $-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$. Let the line $L$ passing through the points $3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ be a normal to the plane $\pi_2$. If the angle between the planes $\pi_1$ and $\pi_2$ is $\theta$, then $\cos \theta=$