Let $\pi_1$ be the plane determined by the vectors $\hat{i}+\hat{j}$ and $\hat{j}+\hat{k}$,and $\pi_2$ be the plane determined by the vectors $\hat{i}-\hat{j}$ and $\hat{i}+\hat{j}-\hat{k}$. Let $\vec{a}$ be a vector parallel to the line of intersection of $\pi_1$ and $\pi_2$. If $|\vec{a}|=\sqrt{14}$,then $|\vec{a} \cdot(\hat{i}+\hat{j}+\hat{k})|=$

If $L$ is a line common to the planes $3x + 4y + 7z = 1$ and $x - y + z = 5$, then the direction ratios of the line $L$ are:

If the lines $L_1: x = -1 + s, y = 3 - \lambda s, z = 1 + \lambda s$ and $L_2: x = \frac{t}{2}, y = 1 + t, z = 2 - t$ with parameters $s$ and $t$ are coplanar, then $\lambda =$ ?

Find the distance of the point whose position vector is $(2 \hat{i}+\hat{j}-\hat{k})$ from the plane $\vec{r} \cdot(\hat{i}-2 \hat{j}+4 \hat{k})=9$.

Let $\lambda_1, \lambda_2$ be the values of $\lambda$ for which the points $\left(\frac{5}{2}, 1, \lambda\right)$ and $(-2, 0, 1)$ are at equal distance from the plane $2x + 3y - 6z + 7 = 0$. If $\lambda_1 > \lambda_2$,then the distance of the point $(\lambda_1 - \lambda_2, \lambda_2, \lambda_1)$ from the line $\frac{x - 5}{1} = \frac{y - 1}{2} = \frac{z + 7}{2}$ is:

If $P$,$Q$,and $R$ are the feet of the perpendiculars drawn from the point $A(1, 1, 1)$ to the planes $P_1: x + 2y + 2z = 2$,$P_2: 2x - 2y + z = -8$,and to the line of intersection of $P_1$ and $P_2$ respectively,then the area of $\Delta PQR$ is: