Let a straight line $L$ pass through the point $P(2,-1{,}3)$ and be perpendicular to the lines $\frac{x-1}{4}=\frac{y+1}{4}=\frac{z-3}{2}$ and $\frac{x-3}{4}=\frac{y-2}{3}=\frac{z+2}{4}$. If the line $L$ intersects the $yz$ -plane at the point $Q$, then the distance between the points $P$ and $Q$ is:

Step 1: Find the direction vector of line L

The line $L$ is perpendicular to two given lines. The direction vectors of the given lines are $\vec{d}_{1} = \langle 4, 4, 2 \rangle$ and $\vec{d}_{2} = \langle 4, 3, 4 \rangle$. The direction vector of line $L$, denoted as $\vec{d_L}$, must be perpendicular to both $\vec{d}_{1}$ and $\vec{d}_{2}$. Therefore, $\vec{d_L}$ can be found by taking the cross product of $\vec{d}_{1}$ and $\vec{d}_{2}$.

Step 2: Calculate the cross product

We calculate the determinant to find the components of the cross product. This gives us the direction vector $\vec{d_L} = \langle 10, -8, -4 \rangle$. We can simplify this direction vector by dividing by a common factor of 2, so $\vec{d_L} = \langle 5, -4, -2 \rangle$.

Step 3: Write the equation of line L

The line $L$ passes through the point $P(2, -1, 3)$ and has the direction vector $\vec{d_L} = \langle 5, -4, -2 \rangle$. We can write the symmetric form of the equation of line $L$ using these values. We introduce a parameter $\lambda$ to represent any point on the line.

Step 4: Find the point Q where L intersects the yz-plane

The $yz$ -plane is defined by $x=0$. We substitute $x=0$ into the parametric equations of line $L$ to find the value of $\lambda$. Once $\lambda$ is found, we substitute it back into the parametric equations to get the coordinates of point $Q$.

Step 5: Calculate the distance between P and Q

Finally, we use the distance formula to find the distance between point $P(2, -1, 3)$ and point $Q\left(0, \frac{3}{5}, \frac{19}{5}\right)$. We substitute the coordinates into the formula and simplify the expression to get the final distance.

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