Question

If the distance between the plane $Ax - 2y + z = d$ and the plane containing the lines $\dfrac{{x - 1}}{2} = \dfrac{{y - 2}}{3} = \dfrac{{z - 3}}{4}$ and $\dfrac{{x - 2}}{3} = \dfrac{{y - 3}}{4} = \dfrac{{z - 4}}{5}$ is $\sqrt 6$, then the value of $\left| d \right|$ is
A. 3
B. 4
C. 5
D. 6

Answer 1

First of all, find the equation of the plane in which the given two lines are containing. Then use the formula that the distance between the two planes ${a_1}x + {b_1}y + {c_1}z = {d_1}$ and ${a_2}x + {b_2}y + {c_2}z = {d_2}$ is given by $\dfrac{{\left| {{d_2} - {d_1}} \right|}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} }}$ to get the required answer.

Complete step-by-step answer :
Let the given plane equation is ${P_1}:Ax - 2y + z = d$
Let ${P_2}$ be the equation of the plane containing the lines $\dfrac{{x - 1}}{2} = \dfrac{{y - 2}}{3} = \dfrac{{z - 3}}{4}$ and $\dfrac{{x - 2}}{3} = \dfrac{{y - 3}}{4} = \dfrac{{z - 4}}{5}$.
We know that the plane equation containing the lines $\dfrac{{x - {x_1}}}{{{p_1}}} = \dfrac{{y - {y_1}}}{{{p_2}}} = \dfrac{{z - {z_1}}}{{{p_3}}}$ and $\dfrac{{x - {x_2}}}{{{q_1}}} = \dfrac{{y - {y_2}}}{{{q_2}}} = \dfrac{{z - {z_2}}}{{{q_3}}}$ is given by \left| {\begin{array}{*{20}{c}} {x - {x_1}}&{y - {y_1}}&{z - {z_1}} \\\ {{p_1}}&{{p_2}}&{{p_3}} \\\ {{q_1}}&{{q_2}}&{{q_3}} \end{array}} \right| = 0 or \left| {\begin{array}{*{20}{c}} {x - {x_2}}&{y - {y_2}}&{z - {z_2}} \\\ {{p_1}}&{{p_2}}&{{p_3}} \\\ {{q_1}}&{{q_2}}&{{q_3}} \end{array}} \right| = 0.
So, the equation of plane ${P_2}$ is given by

{x - 1}&{y - 2}&{z - 3} \\\ 2&3&4 \\\ 3&4&5 \end{array}} \right| = 0$$ Opening the determinant, we have

\Rightarrow \left( {x - 1} \right)\left[ {3 \times 5 - 4 \times 4} \right] - \left( {y - 2} \right)\left[ {2 \times 5 - 3 \times 4} \right] + \left( {z - 3} \right)\left[ {2 \times 4 - 3 \times 3} \right] = \\
\Rightarrow \left( {x - 1} \right)\left( {15 - 16} \right) - \left( {y - 2} \right)\left( {10 - 12} \right) + \left( {z - 3} \right)\left( {8 - 9} \right) = 0 \\
\Rightarrow - \left( {x - 1} \right) + 2\left( {y - 2} \right) - \left( {z - 3} \right) = 0 \\
\Rightarrow - x + 1 + 2y - 4 - z + 3 = 0 \\
\therefore {P_2}:x - 2y + z = 0 \\

We know that the distance between the two planes $${a_1}x + {b_1}y + {c_1}z = {d_1}$$ and $${a_2}x + {b_2}y + {c_2}z = {d_2}$$ is given by $$\dfrac{{\left| {{d_2} - {d_1}} \right|}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} }}$$. Given that the distance between the two planes $${P_1}\& {P_2}$$ is $$\sqrt 6 $$. So, we have

\Rightarrow \sqrt 6 = \dfrac{{\left| {d - 0} \right|}}{{\sqrt {{1^2} + {{\left( { - 2} \right)}^2} + {1^2}} }} \\
\Rightarrow \sqrt 6 = \dfrac{{\left| d \right|}}{{\sqrt {1 + 4 + 1} }} \\
\Rightarrow \sqrt 6 \times \sqrt 6 = \left| d \right| \\
\therefore \left| d \right| = 6 \\

Thus, the correct option is D. 6 **Note** : The plane equation containing the lines $$\dfrac{{x - {x_1}}}{{{p_1}}} = \dfrac{{y - {y_1}}}{{{p_2}}} = \dfrac{{z - {z_1}}}{{{p_3}}}$$ and $$\dfrac{{x - {x_2}}}{{{q_1}}} = \dfrac{{y - {y_2}}}{{{q_2}}} = \dfrac{{z - {z_2}}}{{{q_3}}}$$ is given by $$\left| {\begin{array}{*{20}{c}} {x - {x_1}}&{y - {y_1}}&{z - {z_1}} \\\ {{p_1}}&{{p_2}}&{{p_3}} \\\ {{q_1}}&{{q_2}}&{{q_3}} \end{array}} \right| = 0$$ or $$\left| {\begin{array}{*{20}{c}} {x - {x_2}}&{y - {y_2}}&{z - {z_2}} \\\ {{p_1}}&{{p_2}}&{{p_3}} \\\ {{q_1}}&{{q_2}}&{{q_3}} \end{array}} \right| = 0$$.