Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}

Question

If the distances of the point $ (1,2,a) $ from the line \[ \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1} \] along the lines \[ L_1:\ \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b} \quad \text{and} \quad L_2:\ \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c} \] are equal, then $ a+b+c $ is equal to:

Answer 1

To find the value of $27(QR)^2$, we first need to determine the coordinates of the intersection point $R$ of the lines $L_1$ and $L_2$. After finding $R$, we'll find the specific points $P$ on $L_1$ and $Q$ on $L_2$ using the given conditions. Let's start with the equations of the lines:

The line $L_1$ is given by:

\[ \vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k}), \, \lambda \in \mathbb{R} \] Which gives the parametric equations:
\[ x = 1 + 2\lambda, \quad y = 2 + 3\lambda, \quad z = 3 + 4\lambda \]

The line $L_2$ is given by:

\[ \vec{r} = (4\hat{i} + \hat{j}) + \mu(5\hat{i} + 2\hat{j} + \hat{k}), \, \mu \in \mathbb{R} \] Which gives the parametric equations:
\[ x = 4 + 5\mu, \quad y = 1 + 2\mu, \quad z = \mu \]

To find intersecting point $R$, equate the parametric equations:

1. $1 + 2\lambda = 4 + 5\mu$

2. $2 + 3\lambda = 1 + 2\mu$

3. $3 + 4\lambda = \mu$

Solving these equations simultaneously:

From equation (3), we get:

$\mu = 3 + 4\lambda$

Substitute $\mu = 3 + 4\lambda$ in equation (1):

\[ 1 + 2\lambda = 4 + 5(3 + 4\lambda) \] \[ 1 + 2\lambda = 4 + 15 + 20\lambda \] \[ 1 + 2\lambda = 19 + 20\lambda \] \[ 18\lambda = -18 \] \[ \lambda = -1 \]

Substitute $\lambda = -1$ back into the equation for $\mu$:

$\mu = 3 + 4(-1) = -1$

The coordinates of the intersection point $R$ are:

Substitute $\lambda = -1$ into line $L_1$

\[ x = 1 + 2(-1) = -1, \quad y = 2 + 3(-1) = -1, \quad z = 3 + 4(-1) = -1 \]

The intersection point $R(-1, -1, -1)$.

Now, find the point $P$ on line $L_1$ such that $|PR| = \sqrt{29}$:

Let $P$ be $(1 + 2\alpha, 2 + 3\alpha, 3 + 4\alpha)$, and since $|PR| = \sqrt{29}$, we solve:

\[ \sqrt{(1 + 2\alpha + 1)^2 + (2 + 3\alpha + 1)^2 + (3 + 4\alpha + 1)^2} = \sqrt{29} \]

Simplifying:

\[ (2\alpha + 2)^2 + (3\alpha + 3)^2 + (4\alpha + 4)^2 = 29 \]

Solving this gives $\alpha = 1$, which means point $P$ is $(3, 5, 7)$ (in the first octant).

Now, find the point $Q$ on line $L_2$ such that $|PQ| = \sqrt{\frac{47}{3}}$:

Let $Q$ be $(4 + 5\beta, 1 + 2\beta, \beta)$, where $|PQ| = \sqrt{\frac{47}{3}}$.

Solving $(5\beta + 1)^2 + (2\beta - 4)^2 + (\beta - 7)^2 = \frac{47}{3}$.

Through this computation, we obtain $\beta = \frac{1}{3}$.

Now, locate point $Q$: $Q = (4 + 5 \frac{1}{3}, 1 + 2 \frac{1}{3}, \frac{1}{3}) = (5 \frac{2}{3}, 1 \frac{2}{3}, \frac{1}{3})$.

Finally, compute $27(QR)^2$:

Coordinates of $Q = \left(\frac{17}{3}, \frac{5}{3}, \frac{1}{3}\right)$.

Calculate $QR$ as $\sqrt{\left(-1 - \frac{17}{3}\right)^2 + \left(-1 - \frac{5}{3}\right)^2 + \left(-1 - \frac{1}{3}\right)^2}$ :

And $QR^2 = \frac{348}{27}$. Hence, $27 (QR)^2 = 348$.

Thus, the answer is 348.

Answer 2