If the equation of the line passing through the point $ \left( 0, -\frac{1}{2}, 0 \right) $ and perpendicular to the lines $ \vec{r} = \lambda \left( \hat{i} + a\hat{j} + b\hat{k} \right) $ and $ \vec{r} = \left( \hat{i} - \hat{j} - 6\hat{k} \right) + \mu \left( -b \hat{i} + a\hat{j} + 5\hat{k} \right) $ is $ \frac{x-1}{-2} = \frac{y+4}{d} = \frac{z-c}{-4} $, then $ a+b+c+d $ is equal to :

<p>The line we need to find should be orthogonal to two given lines, meaning it will be parallel to the cross product of their direction vectors: </p> <p>$ (\hat{i} + a \hat{j} + b \hat{k}) \times (-b \hat{i} + a \hat{j} + 5 \hat{k}) $</p> <p>Calculating the cross product results in the vector:</p> <p>$ \hat{i}(5a - ab) - \hat{j}(b^2 + 5) + \hat{k}(a + ab) $</p> <p>This shows that the direction ratios of the required line are proportional to:</p> <p>$ \alpha(5a - ab), - (b^2 + 5), (a + ab) $</p> <p>Since the line is also expressed as $ \frac{x-1}{-2} = \frac{y+4}{d} = \frac{z-c}{-4} $, the direction ratios based on this equation are:</p> <p>$ \alpha(-2), d, -4 $</p> <p>Thus, we have the equation:</p> <p>$ \frac{5a - ab}{-2} = \frac{- (b^2 + 5)}{d} = \frac{a + ab}{-4} $</p> <p>The point $\left(0, -\frac{1}{2}, 0 \right)$ lies on the line, which implies:</p> <p>$ \frac{0-1}{-2} = \frac{-\frac{1}{2} + 4}{d} = \frac{0-c}{-4} $</p> <p>Simplifying these gives $ d = 7 $ and $ c = 2 $.</p> <p>Substitute into the main equation:</p> <p>$ \frac{5a - ab}{-2} = \frac{a + ab}{-4} $</p> <p>$ \frac{-b^2 - 5}{7} = \frac{a + ab}{-4} $</p> <p>Solving these yields:</p> <p>$ \begin{array}{c|c} -20a + 4ab = -2a - 2ab & 4b^2 + 20 = 70 + 7ab \\ 18a = 6ab & 36 + 20 = 70 + 21a \\ b = 3 & 56 = 28a \Rightarrow a = 2 \end{array} $</p> <p>Substituting these into the variables, we find:</p> <p>$ a + b + c + d = 2 + 3 + 2 + 7 = 14 $</p>